The first step is to write a balanced chemical equation for this reaction:

 2AgNO₃(aq) + MgBr₂(aq) → 2AgBr(s) + Mg(NO₃)₂(aq)

The plan is to calculate the moles of AgNO₃, then determine the moles of MgBr₂ based on the stoichiometric ratio, and finally convert the moles of MgBr₂ to mL using the formula for molarity.

 To calculate the moles, we need to use the formula for molarity in which the volume must be in liters because the unit for molarity is mol/L. . So, convert mL to L first:

\[{\rm{V}}\left( {{\rm{AgN}}{{\rm{O}}_{\rm{3}}}} \right)\;{\rm{ = }}\;{\rm{600}}{\rm{.0}}\;\cancel{{{\rm{mL}}}}\;{\rm{ \times }}\;\frac{{{\rm{1}}\;{\rm{L}}}}{{{\rm{1000}}\;\cancel{{{\rm{mL}}}}}}\;{\rm{ = }}\;{\rm{0}}{\rm{.6000}}\;{\rm{L}}\]

\[{\rm{M}}\;{\rm{ = }}\;\frac{{\rm{n}}}{{\rm{V}}}\]

 n = MV

\[{\rm{n}}\left( {{\rm{AgN}}{{\rm{O}}_{\rm{3}}}} \right)\;{\rm{ = }}\;{\rm{0}}{\rm{.80}}\;\frac{{{\rm{mol}}}}{{\cancel{{\rm{L}}}}}\;{\rm{ \times }}\;{\rm{0}}{\rm{.6000}}\;\cancel{{\rm{L}}}\;{\rm{ = }}\;{\rm{0}}{\rm{.48}}\;{\rm{mol}}\]

Now, we use this to determine the moles of MgBr₂ based on the stoichiometric ratio:

\[{\rm{n}}\left( {{\rm{MgB}}{{\rm{r}}_{\rm{2}}}} \right)\;{\rm{ = }}\;{\rm{0}}{\rm{.48}}\;\cancel{{{\rm{mol}}\;{\rm{AgN}}{{\rm{O}}_{\rm{3}}}}}\;{\rm{ \times }}\;\frac{{{\rm{1}}\;{\rm{mol}}\;{\rm{MgB}}{{\rm{r}}_{\rm{2}}}}}{{{\rm{2}}\;\cancel{{{\rm{mol}}\;{\rm{AgN}}{{\rm{O}}_{\rm{3}}}}}}}\;{\rm{ = }}\;{\rm{0}}{\rm{.24}}\;{\rm{mol}}\]

\[{\rm{V}}\left( {{\rm{MgB}}{{\rm{r}}_{\rm{2}}}} \right)\;{\rm{ = }}\;\frac{{\rm{n}}}{{\rm{M}}}\;{\rm{ = }}\;\frac{{{\rm{0}}{\rm{.24}}\;{\rm{mol}}}}{{{\rm{1}}{\rm{.80}}\;{\rm{mol/L}}}}{\rm{ = }}\;{\rm{0}}{\rm{.13}}\;{\rm{L}}\]

To convert this to mL, we multiply it by 1000, so it is 130 mL.

The first step is to write a balanced chemical equation for this reaction: 

K₂SO₄ (aq) + BaCl₂ (aq) → BaSO₄ (s) + 2KCl(aq)

The plan is to calculate the moles of BaCl₂, then determine the moles of K₂SO₄ based on the stoichiometric ratio, and finally convert the moles of K₂SO₄ to mL using the formula for molarity.

 To calculate the moles, we need to use the formula for molarity in which the volume must be in liters because the unit for molarity is mol/L. So, convert mL to L first:

\[{\rm{V}}\left( {{\rm{BaC}}{{\rm{l}}_{\rm{2}}}} \right)\;{\rm{ = }}\;5{\rm{00}}{\rm{.}}\;\cancel{{{\rm{mL}}}}\;{\rm{ \times }}\;\frac{{{\rm{1}}\;{\rm{L}}}}{{{\rm{1000}}\;\cancel{{{\rm{mL}}}}}}\;{\rm{ = }}\;{\rm{0}}{\rm{.500}}\;{\rm{L}}\]

\[{\rm{M}}\;{\rm{ = }}\;\frac{{\rm{n}}}{{\rm{V}}}\]

n = MV

\[{\rm{n}}\left( {{\rm{BaC}}{{\rm{l}}_{\rm{2}}}} \right)\;{\rm{ = }}\;{\rm{0}}{\rm{.650}}\;\frac{{{\rm{mol}}}}{{\cancel{{\rm{L}}}}}\;{\rm{ \times }}\;{\rm{0}}{\rm{.500}}\;\cancel{{\rm{L}}}\;{\rm{ = }}\;{\rm{0}}{\rm{.325}}\;{\rm{mol}}\]

Now, we use this to determine the moles of K₂SO₄ based on the stoichiometric ratio:

\[{\rm{n}}\left( {{{\rm{K}}_{\rm{2}}}{\rm{S}}{{\rm{O}}_{\rm{4}}}} \right)\;{\rm{ = }}\;{\rm{0}}{\rm{.325}}\;\cancel{{{\rm{mol}}\;{\rm{BaC}}{{\rm{l}}_{\rm{2}}}}}\;{\rm{ \times }}\;\frac{{{\rm{1}}\;{\rm{mol}}\;{{\rm{K}}_{\rm{2}}}{\rm{S}}{{\rm{O}}_{\rm{4}}}}}{{{\rm{1}}\;\cancel{{{\rm{mol}}\;{\rm{BaC}}{{\rm{l}}_{\rm{2}}}}}}}\;{\rm{ = }}\;{\rm{0}}{\rm{.325}}\;{\rm{mol}}\]

Next, use the moles in the formula for molarity and determine the volume of the solution in L:

\[{\rm{V}}\left( {{\rm{MgB}}{{\rm{r}}_{\rm{2}}}} \right)\;{\rm{ = }}\;\frac{{\rm{n}}}{{\rm{M}}}\;{\rm{ = }}\;\frac{{{\rm{0}}{\rm{.325}}\;{\rm{mol}}}}{{{\rm{2}}{\rm{.70}}\;{\rm{mol/L}}}}{\rm{ = }}\;{\rm{0}}{\rm{.120}}\;{\rm{L}}\]

To convert this to mL, we multiply it by 1000, so it is 120 mL.

First, let’s write a balanced chemical equation for this reaction:

NaOH + KHC₈H₄O₄ → NaKC₈H₄O₄ + H₂O

The plan here would be calculating the moles of KHP, using it to determine the moles of NaOH based on the stoichiometric ration, and use that to calculate the volume of NaOH solution:

\[{\rm{n}}\left( {{\rm{KHP}}} \right)\;{\rm{ = }}\;\frac{{\rm{m}}}{{\rm{M}}}\;{\rm{ = }}\;\frac{{{\rm{0}}{\rm{.326}}\;{\rm{g}}}}{{{\rm{204}}{\rm{.2}}\;{\rm{g/mol}}}}{\rm{ = }}\;{\rm{0}}{\rm{.00160}}\;{\rm{mol}}\]

Because the ratio of NaOH and KHP is 1:1, the moles are going to be equal and therefore, there are 0.00160 mol NaOH reacting with the KHP.

Now, we can use the moles and the volume of the solution to determine the molarity of NaOH. Don’t forget to convert mL to L when working with molarity:

\[{\rm{V}}\left( {{\rm{NaOH}}} \right)\;{\rm{ = }}\;27.0\;\cancel{{{\rm{mL}}}}\;{\rm{ \times }}\;\frac{{{\rm{1}}\;{\rm{L}}}}{{{\rm{1000}}\;\cancel{{{\rm{mL}}}}}}\;{\rm{ = }}\;{\rm{0}}{\rm{.0270}}\;{\rm{L}}\]

\[{\rm{M}}\left( {{\rm{NaOH}}} \right)\;{\rm{ = }}\;\frac{{\rm{n}}}{{\rm{V}}}\;{\rm{ = }}\;\frac{{{\rm{0}}{\rm{.00160}}\;{\rm{mol}}}}{{{\rm{0}}{\rm{.0270}}\;{\rm{L}}}}{\rm{ = }}\;{\rm{0}}{\rm{.0593}}\;{\rm{mol/L}}\]

First, let’s write a balanced chemical equation for this reaction:

Ca(OH)₂(s) + H₂SO₄(aq) → CaSO₄(aq) + 2H₂O(l)

The linkage between the two reactants/solutions is still going to be the mole conversion based on the stoichiometric ratio. So, to plan is as follows:

To calculate the moles, we need to use the formula for molarity in which the volume must be in liters because the unit for molarity is mol/L. So, convert mL to L first:

\[{\rm{V}}\left( {{{\rm{H}}_{\rm{2}}}{\rm{S}}{{\rm{O}}_{\rm{4}}}} \right)\;{\rm{ = }}\;{\rm{250}}{\rm{.0}}\;\cancel{{{\rm{mL}}}}\;{\rm{ \times }}\;\frac{{{\rm{1}}\;{\rm{L}}}}{{{\rm{1000}}\;\cancel{{{\rm{mL}}}}}}\;{\rm{ = }}\;{\rm{0}}{\rm{.2500}}\;{\rm{L}}\]

\[{\rm{M}}\;{\rm{ = }}\;\frac{{\rm{n}}}{{\rm{V}}}\]

n = MV

\[{\rm{n}}\left( {{{\rm{H}}_{\rm{2}}}{\rm{S}}{{\rm{O}}_{\rm{4}}}} \right)\;{\rm{ = }}\;{\rm{0}}{\rm{.0350}}\;\frac{{{\rm{mol}}}}{{\cancel{{\rm{L}}}}}\;{\rm{ \times }}\;{\rm{0}}{\rm{.2500}}\;\cancel{{\rm{L}}}\;{\rm{ = }}\;{\rm{0}}{\rm{.00875}}\;{\rm{mol}}\]

Now, we use this to determine the moles of Ca(OH)₂ based on the stoichiometric ratio:

\[{\rm{n}}\left( {{\rm{Ca}}{{\left( {{\rm{OH}}} \right)}_{\rm{2}}}} \right)\;{\rm{ = }}\;{\rm{0}}{\rm{.00875}}\;\cancel{{{\rm{mol}}\;{{\rm{H}}_{\rm{2}}}{\rm{S}}{{\rm{O}}_{\rm{4}}}}}\;{\rm{ \times }}\;\frac{{{\rm{1}}\;{\rm{mol}}\;{\rm{Ca}}{{\left( {{\rm{OH}}} \right)}_{\rm{2}}}}}{{{\rm{1}}\;\cancel{{{\rm{mol}}\;{{\rm{H}}_{\rm{2}}}{\rm{S}}{{\rm{O}}_{\rm{4}}}}}}}\;{\rm{ = }}\;{\rm{0}}{\rm{.00875}}\;{\rm{mol}}\]

Next, use the moles in the formula for molarity and determine the volume of the solution in L:

\[{\rm{V}}\left( {{\rm{Ca}}{{\left( {{\rm{OH}}} \right)}_{\rm{2}}}} \right)\;{\rm{ = }}\;\frac{{\rm{n}}}{{\rm{M}}}\;{\rm{ = }}\;\frac{{{\rm{0}}{\rm{.00875}}\;{\rm{mol}}}}{{{\rm{0}}{\rm{.150}}\;{\rm{mol/L}}}}{\rm{ = }}\;{\rm{0}}{\rm{.0583}}\;{\rm{L}}\]

First, let’s write a balanced chemical equation for this reaction:

Ba(OH)₂(s) + 2HCl(aq) → BaCl₂(aq) + 2H₂O(l)

The linkage between the two reactants/solutions is still going to be the mole conversion based on the stoichiometric ratio. So, the plan is as follows:

To calculate the moles, we need to use the formula for molarity in which the volume must be in liters because the unit for molarity is mol/L. So, convert mL to L first:

\[{\rm{V}}\left( {{\rm{Ba}}{{\left( {{\rm{OH}}} \right)}_{\rm{2}}}} \right)\;{\rm{ = }}\;{\rm{60}}{\rm{.00}}\;\cancel{{{\rm{mL}}}}\;{\rm{ \times }}\;\frac{{{\rm{1}}\;{\rm{L}}}}{{{\rm{1000}}\;\cancel{{{\rm{mL}}}}}}\;{\rm{ = }}\;{\rm{0}}{\rm{.06000}}\;{\rm{L}}\]

\[{\rm{M}}\;{\rm{ = }}\;\frac{{\rm{n}}}{{\rm{V}}}\]

n = MV

\[{\rm{n}}\left( {{\rm{Ba}}{{\left( {{\rm{OH}}} \right)}_{\rm{2}}}} \right)\;{\rm{ = }}\;{\rm{0}}{\rm{.060}}\;\frac{{{\rm{mol}}}}{{\cancel{{\rm{L}}}}}\;{\rm{ \times }}\;{\rm{0}}{\rm{.06000}}\;\cancel{{\rm{L}}}\;{\rm{ = }}\;{\rm{0}}{\rm{.0036}}\;{\rm{mol}}\]

Now, we use this to determine the moles of HCl based on the stoichiometric ratio:

\[{\rm{n}}\left( {{\rm{HCl}}} \right)\;{\rm{ = }}\;{\rm{0}}{\rm{.0036}}\;\cancel{{{\rm{mol}}\;{\rm{Ba}}{{\left( {{\rm{OH}}} \right)}_{\rm{2}}}}}\;{\rm{ \times }}\;\frac{{{\rm{2}}\;{\rm{mol}}\;{\rm{HCl}}}}{{{\rm{1}}\;\cancel{{{\rm{mol}}\;{\rm{Ba}}{{\left( {{\rm{OH}}} \right)}_{\rm{2}}}}}}}\;{\rm{ = }}\;{\rm{0}}{\rm{.0072}}\;{\rm{mol}}\]

Next, use the moles in the formula for molarity and determine the volume of the solution in L and convert it to mL:

\[{\rm{V}}\left( {{\rm{HCl}}} \right)\;{\rm{ = }}\;\frac{{\rm{n}}}{{\rm{M}}}\;{\rm{ = }}\;\frac{{{\rm{0}}{\rm{.0072}}\;{\rm{mol}}}}{{{\rm{0}}{\rm{.250}}\;{\rm{mol/L}}}}{\rm{ = }}\;{\rm{0}}{\rm{.0288}}\;{\rm{L}}\]

To get the mL, we just need to multiply by 1000, or use the molar conversion:

\[{\rm{V}}\left( {{\rm{HCl}}} \right)\;{\rm{ = }}\;{\rm{0}}{\rm{.0288}}\;{\rm{L}}\;{\rm{ \times }}\;\frac{{{\rm{1000}}\;{\rm{mL}}}}{{{\rm{1}}\;{\rm{L}}}}\;{\rm{ = }}\;{\rm{28}}{\rm{.8}}\;{\rm{mL}}\]

The first step is to write a balanced chemical equation for this reaction:

2AgNO₃(aq) + MgCl₂(aq) → 2AgCl(s) + Mg(NO₃)₂(aq)

Next, calculate the moles and determine the limiting reactant. For this, we need to use the formula for molarity:

n = MV

Before entering the numbers, remember to convert the volume to liters because the unit for molarity is mol/L.

\[{\rm{V}}\left( {{\rm{AgN}}{{\rm{O}}_{\rm{3}}}} \right)\;{\rm{ = }}\;{\rm{20}}{\rm{.0}}\;\cancel{{{\rm{mL}}}}\;{\rm{ \times }}\;\frac{{{\rm{1}}\;{\rm{L}}}}{{{\rm{1000}}\;\cancel{{{\rm{mL}}}}}}\;{\rm{ = }}\;{\rm{0}}{\rm{.0200}}\;{\rm{L}}\]

\[{\rm{V}}\left( {{\rm{MgC}}{{\rm{l}}_{\rm{2}}}} \right)\;{\rm{ = }}\;{\rm{45}}{\rm{.0}}\;\cancel{{{\rm{mL}}}}\;{\rm{ \times }}\;\frac{{{\rm{1}}\;{\rm{L}}}}{{{\rm{1000}}\;\cancel{{{\rm{mL}}}}}}\;{\rm{ = }}\;{\rm{0}}{\rm{.0450}}\;{\rm{L}}\]

And now, we can use the volumes and concentrations to calculate the moles:

\[{\rm{n}}\left( {{\rm{AgN}}{{\rm{O}}_{\rm{3}}}} \right)\;{\rm{ = }}\;{\rm{0}}{\rm{.150}}\;\frac{{{\rm{mol}}}}{{\cancel{{\rm{L}}}}}\;{\rm{ \times }}\;{\rm{0}}{\rm{.0200}}\;\cancel{{\rm{L}}}\;{\rm{ = }}\;{\rm{0}}{\rm{.00300}}\;{\rm{mol}}\]

\[{\rm{n}}\left( {{\rm{MgC}}{{\rm{l}}_{\rm{2}}}} \right)\;{\rm{ = }}\;{\rm{0}}{\rm{.250}}\;\frac{{{\rm{mol}}}}{{\cancel{{\rm{L}}}}}\;{\rm{ \times }}\;{\rm{0}}{\rm{.0450}}\;\cancel{{\rm{L}}}\;{\rm{ = }}\;{\rm{0}}{\rm{.0113}}\;{\rm{mol}}\]

Remember, limiting reactant is the one that gives less product. We can do the calculations base don any product, but since we need to determine the mass of AgCl produced, let’s do the calculations based on it:

\[{\rm{n}}\left( {{\rm{AgCl}}} \right)\;{\rm{ = }}\;{\rm{0}}{\rm{.00300}}\;\cancel{{{\rm{mol}}\;{\rm{AgN}}{{\rm{O}}_{\rm{3}}}}}\;{\rm{ \times }}\;\frac{{{\rm{2}}\;{\rm{mol}}\;{\rm{AgCl}}}}{{{\rm{2}}\;\cancel{{{\rm{mol}}\;{\rm{AgN}}{{\rm{O}}_{\rm{3}}}}}}}\;{\rm{ = }}\;{\rm{0}}{\rm{.00300}}\;{\rm{mol}}\]

\[{\rm{n}}\left( {{\rm{AgCl}}} \right)\;{\rm{ = }}\;{\rm{0}}{\rm{.0113 }}\cancel{{{\rm{mol}}\;{\rm{MgC}}{{\rm{l}}_{\rm{2}}}}}\;{\rm{ \times }}\;\frac{{{\rm{2}}\;{\rm{mol}}\;{\rm{AgCl}}}}{{{\rm{1}}\;\cancel{{{\rm{mol}}\;{\rm{MgC}}{{\rm{l}}_{\rm{2}}}}}}}\;{\rm{ = }}\;{\rm{0}}{\rm{.0226}}\;{\rm{mol}}\]

AgNO₃ produces less product and therefore, it is the limiting reactant, and we must do our calculations based on it.

The last step is to convert the moles of AgCl to mass:

\[{\rm{m}}\left( {{\rm{AgCl}}} \right)\;{\rm{ = }}\;{\rm{0}}{\rm{.00300}}\;\cancel{{{\rm{mol}}}}\;{\rm{ \times }}\;{\rm{143}}{\rm{.3}}\;\frac{{\;{\rm{g}}}}{{\cancel{{{\rm{mol}}}}}}\;{\rm{ = }}\;{\rm{0}}{\rm{.430}}\;{\rm{g}}\]

First, write a balanced chemical equation for this reaction:

K₂SO₄(aq) + BaCl₂(aq) → BaSO₄(s) + 2KCl(aq)

The amount of product formed is given, so we know the actual yield of the reaction. This means our calculations for the amounts of reactants that reacted must be done based on the moles of the product. The precipitate is BaSO_4, and to determine its moles, we use the formula relating the mass, molar mass, and the moles:

\[{\rm{n(BaS}}{{\rm{O}}_{\rm{4}}}{\rm{)}}\;{\rm{ = }}\;\frac{{{\rm{31}}{\rm{.2 g}}}}{{{\rm{233}}{\rm{.4}}\;{\rm{g/mol}}}}\;{\rm{ = }}\;{\rm{0}}{\rm{.134}}\;{\rm{mol}}\;\]

This already is enough to calculate the volume of the K₂SO₄ solution because we can determine the moles of K₂SO₄ based on the stoichiometric ratio. Every one mole of K₂SO₄ produces one mole of BaSO₄ and therefore, 0.134 mol K₂SO₄ had reacted.

Next, use the moles in the formula for molarity and determine the volume of the solution in L and convert it to mL:

\[{\rm{V}}\left( {{{\rm{K}}_{\rm{2}}}{\rm{S}}{{\rm{O}}_{\rm{4}}}} \right)\;{\rm{ = }}\;\frac{{\rm{n}}}{{\rm{M}}}\;{\rm{ = }}\;\frac{{{\rm{0}}{\rm{.134}}\;{\rm{mol}}}}{{{\rm{0}}{\rm{.600}}\;{\rm{mol/L}}}}{\rm{ = }}\;{\rm{0}}{\rm{.223}}\;{\rm{L}}\]

To get the mL, we just need to multiply by 1000, or use the molar conversion:

\[{\rm{V}}\left( {{{\rm{K}}_{\rm{2}}}{\rm{S}}{{\rm{O}}_{\rm{4}}}} \right)\;{\rm{ = }}\;{\rm{0}}{\rm{.223}}\;{\rm{L}}\;{\rm{ \times }}\;\frac{{{\rm{1000}}\;{\rm{mL}}}}{{{\rm{1}}\;{\rm{L}}}}\;{\rm{ = }}\;{\rm{223}}\;{\rm{mL}}\]

To confirm that BaCl₂ was un excess, we determine its moles from the concentration and volume which is 0.4000 L (400.0 mL):

\[{\rm{n(BaC}}{{\rm{l}}_{\rm{2}}}{\rm{)}}\;{\rm{ = }}\;{\rm{0}}{\rm{.500}}\;\frac{{{\rm{mol}}}}{{\cancel{{\rm{L}}}}}\;{\rm{ \times }}\;{\rm{0}}{\rm{.4000}}\;\cancel{{\rm{L}}}\;{\rm{ = }}\;{\rm{0}}{\rm{.200}}\;{\rm{mol}}\]

Now, 0.200 mol of BaCl₂ would have produced as much of BaSO₄ since they have a 1:1 molar ratio. This amount is greater than what was formed from 0.134 mol K₂SO₄ which proves that BaCl₂ was present in excess and K₂SO₄ was the limiting reactant.

The quantities of both reactants are given, so we need to determine the limiting reactant. Remember, limiting reactant is the one that gives less product. We can do the calculations based on any product, but since we are working in PbCl₂, let’s do the calculations based on it.

The first step in doing these calculations, if to find the moles of both reactants for which we use the formula for molarity. As always, convert the mL to L first:

\[{\rm{V}}\left( {{\rm{NaCl}}} \right)\;{\rm{ = }}\;{\rm{42}}{\rm{.0}}\;{\rm{mL}}\;{\rm{ \times }}\;\frac{{{\rm{1}}\;{\rm{L}}}}{{{\rm{1000}}\;{\rm{mL}}}}\;{\rm{ = }}\;{\rm{0}}{\rm{.042}}\;{\rm{L}}\]

\[{\rm{V}}\left( {{\rm{Pb}}{{\left( {{\rm{N}}{{\rm{O}}_{\rm{3}}}} \right)}_{\rm{2}}}} \right)\;{\rm{ = }}\;{\rm{17}}{\rm{.0}}\;{\rm{mL}}\;{\rm{ \times }}\;\frac{{{\rm{1}}\;{\rm{L}}}}{{{\rm{1000}}\;{\rm{mL}}}}\;{\rm{ = }}\;{\rm{0}}{\rm{.017}}\;{\rm{L}}\]

\[{\rm{n(NaCl)}}\;{\rm{ = }}\;{\rm{0}}{\rm{.85}}\;\frac{{{\rm{mol}}}}{{\cancel{{\rm{L}}}}}\;{\rm{ \times }}\;{\rm{0}}{\rm{.0420}}\;\cancel{{\rm{L}}}\;{\rm{ = }}\;{\rm{0}}{\rm{.0357}}\;{\rm{mol}}\]

\[{\rm{n(Pb}}{\left( {{\rm{N}}{{\rm{O}}_{\rm{3}}}} \right)_{\rm{2}}}{\rm{)}}\;{\rm{ = }}\;{\rm{0}}{\rm{.650}}\;\frac{{{\rm{mol}}}}{{\cancel{{\rm{L}}}}}\;{\rm{ \times }}\;{\rm{0}}{\rm{.0170}}\;\cancel{{\rm{L}}}\;{\rm{ = }}\;{\rm{0}}{\rm{.0111}}\;{\rm{mol}}\]

Now, let’s see which of these produces less PbCl_2:

\[{\rm{n}}\left( {{\rm{PbC}}{{\rm{l}}_{\rm{2}}}} \right)\;{\rm{ = }}\;{\rm{0}}{\rm{.0357}}\;\cancel{{{\rm{mol}}\;{\rm{NaCl}}}}\;{\rm{ \times }}\;\frac{{{\rm{1}}\;{\rm{mol}}\;{\rm{PbC}}{{\rm{l}}_{\rm{2}}}}}{{{\rm{2}}\;\cancel{{{\rm{mol}}\;{\rm{NaCl}}}}}}\;{\rm{ = }}\;{\rm{0}}{\rm{.0179}}\;{\rm{mol}}\]

\[{\rm{n}}\left( {{\rm{PbC}}{{\rm{l}}_{\rm{2}}}} \right)\;{\rm{ = }}\;{\rm{0}}{\rm{.0111}}\;\cancel{{{\rm{mol}}\;{\rm{Pb}}{{\left( {{\rm{N}}{{\rm{O}}_{\rm{3}}}} \right)}_{\rm{2}}}}}\;{\rm{ \times }}\;\frac{{{\rm{1}}\;{\rm{mol}}\;{\rm{PbC}}{{\rm{l}}_{\rm{2}}}}}{{{\rm{1}}\;\cancel{{{\rm{mol}}\;{\rm{Pb}}{{\left( {{\rm{N}}{{\rm{O}}_{\rm{3}}}} \right)}_{\rm{2}}}}}}}\;{\rm{ = }}\;{\rm{0}}{\rm{.0111}}\;{\rm{mol}}\]

Pb(NO₃)₂ is the limiting reactant because it gives less product.

The theoretical yield of the reaction, therefore, is 0.0111 mol which is:

\[{\rm{m}}\left( {{\rm{PbC}}{{\rm{l}}_{\rm{2}}}} \right)\;{\rm{ = }}\;{\rm{0}}{\rm{.0111}}\;\cancel{{{\rm{mol}}}}\;{\rm{ \times }}\;\frac{{{\rm{278}}{\rm{.1}}\;{\rm{g}}}}{{\cancel{{{\rm{mol}}}}}}\;{\rm{ = }}\;{\rm{3}}{\rm{.09}}\;{\rm{g}}\]

The percent yield of the reaction is the ratio of the actual over the theoretical yield:

\[{\rm{\% }}\;{\rm{Yield}}\;{\rm{ = }}\;\frac{{\rm{A}}}{{\rm{T}}}\;{\rm{ = }}\;\frac{{{\rm{2}}{\rm{.45}}\;{\rm{g}}}}{{{\rm{3}}{\rm{.09}}\;{\rm{g}}}}\; \times \;100\% \; = \;79.3\;\% \]

The plan here would be to find the moles of acetylsalicylic acid based on the amount of NaOH, and use the moles together with the given mass (4.598 g) to determine the molar mass:

We don’t know the formula of acetylsalicylic acid, but we know that it is a monoprotic acid and therefore, the stoichiometric ratio with NaOH is going to be 1:1. To find the moles of NaOH, we are going to use the formula for molarity so, first, express the volume in L:

\[{\rm{V}}\left( {{\rm{NaOH}}} \right)\;{\rm{ = }}\;{\rm{35}}{\rm{.6}}\;{\rm{mL}}\;{\rm{ \times }}\;\frac{{{\rm{1}}\;{\rm{L}}}}{{{\rm{1000}}\;{\rm{mL}}}}\;{\rm{ = }}\;{\rm{0}}{\rm{.0356}}\;{\rm{L}}\]

\[{\rm{n(NaOH)}}\;{\rm{ = }}\;{\rm{0}}{\rm{.717}}\;\frac{{{\rm{mol}}}}{{\cancel{{\rm{L}}}}}\;{\rm{ \times }}\;{\rm{0}}{\rm{.0356}}\;\cancel{{\rm{L}}}\;{\rm{ = }}\;{\rm{0}}{\rm{.0255}}\;{\rm{mol}}\]

Because the moles of the acid and NaOH are equal, 0.0255 mol of acid must’ve reacted. And at this point we have the moles and the mass of the acid. Use these to determine the molar mass:

\[{\rm{M}}\left( {{\rm{acid}}} \right)\;{\rm{ = }}\;\frac{{\rm{m}}}{{\rm{n}}}\;{\rm{ = }}\;\frac{{{\rm{4}}{\rm{.598}}\;{\rm{g}}}}{{{\rm{0}}{\rm{.0255}}\;{\rm{mol}}}}{\rm{ = }}\;{\rm{180}}{\rm{.3}}\;g/{\rm{mol}}\]

The only precipitate that can be formed is AgCl when AgNO₃ reacts with the CaCl_2. So, if we write a balanced chemical equation, we can determine the moles and the mass of the CaCl₂ in the mixture. Once we know the mass of CaCl₂, we subtract it from the total mass of the mixture and get the mass of Ca(NO₃)_2. This, in turn, is divided by the mass of the mixture to get the percentage of Ca(NO₃)_2.

 Let’s write the chemical equation:

2AgNO₃(aq) + CaCl₂(aq) → 2AgCl(s) + Ca(NO₃)₂(aq)

Now, calculate the moles of AgCl and determine the moles of CaCl₂ based on the stoichiometric ratio:

\[{\rm{n}}\left( {{\rm{AgCl}}} \right)\;{\rm{ = }}\;\frac{{\rm{m}}}{{\rm{M}}}\;{\rm{ = }}\;\frac{{{\rm{2}}{\rm{.40}}\;{\rm{g}}}}{{{\rm{143}}{\rm{.3}}\;{\rm{g/mol}}}}{\rm{ = }}\;{\rm{0}}{\rm{.0167}}\;{\rm{mol}}\]

\[{\rm{n}}\left( {{\rm{CaC}}{{\rm{l}}_{\rm{2}}}} \right)\;{\rm{ = }}\;{\rm{0}}{\rm{.0167}}\;\cancel{{{\rm{mol}}\;{\rm{AgCl}}}}\;{\rm{ \times }}\;\frac{{{\rm{1}}\;{\rm{mol}}\;{\rm{CaC}}{{\rm{l}}_{\rm{2}}}}}{{{\rm{2}}\;\cancel{{{\rm{mol}}\;{\rm{AgCl}}}}}}\;{\rm{ = }}\;{\rm{0}}{\rm{.00835}}\;{\rm{mol}}\]

The mass can be calculated using the moles and the molar mass:

\[{\rm{m}}\left( {{\rm{CaC}}{{\rm{l}}_{\rm{2}}}} \right)\;{\rm{ = }}\;{\rm{0}}{\rm{.00835}}\;\cancel{{{\rm{mol}}}}\;{\rm{ \times }}\;\frac{{{\rm{111}}\;{\rm{g}}}}{{\cancel{{{\rm{mol}}}}}}\;{\rm{ = }}\;{\rm{0}}{\rm{.927}}\;{\rm{g}}\]

m(Ca(NO₃)₂)= 2.40 – 0.927 = 1.473 g

\[{\rm{\% }}\;{\rm{Ca}}{\left( {{\rm{N}}{{\rm{O}}_{\rm{3}}}} \right)_{\rm{2}}}\;{\rm{ = }}\;\frac{{{\rm{1}}{\rm{.473}}\;{\rm{g}}}}{{{\rm{2}}{\rm{.40}}\;{\rm{g}}}}\;{\rm{ \times }}\;{\rm{100\% }}\,{\rm{ = }}\;{\rm{61}}{\rm{.4\% }}\]