These calculations are based on the same principles we learn in stoichiometric calculations. The only difference for most problems is going to be the fact that the quantities are given in molarities rather than in moles. Remember that you need to always calculate the moles for solving reaction stoichiometry problems.
For example,
How many ml of 1.80 M MgBr2 is necessary to precipitate all the silver ions in 600.0 mL of 0.80 M AgNO3 solution?
Write a balanced chemical equation for this reaction and follow the plan for the calculation which is to determine the moles, do the stoichiometric calculations, and convert to volume or any other quantity asked in the problem:
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 the mL to L first:
\[{\rm{V}}\left( {{\rm{AgN}}{{\rm{O}}_{\rm{3}}}} \right)\;{\rm{ = }}\;{\rm{600}}{\rm{.0}}\;\cancel{{{\rm{mL}}}}\;{\rm{ \times }}\;\frac{{{\rm{1}}\;{\rm{L}}}}{{{\rm{1000}}\;\cancel{{{\rm{mL}}}}}}\;{\rm{ = }}\;{\rm{0}}{\rm{.6000}}\;{\rm{L}}\]
Next, we rearrange the formula for molarity to calculte the moles of AgNO3:
\[{\rm{M}}\;{\rm{ = }}\;\frac{{\rm{n}}}{{\rm{V}}}\]
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:
\[{\rm{n}}\left( {{\rm{MgB}}{{\rm{r}}_{\rm{2}}}} \right)\;{\rm{ = }}\;{\rm{0}}{\rm{.48}}\;\cancel{{{\rm{mol}}\;{\rm{AgN}}{{\rm{O}}_{\rm{3}}}}}\;{\rm{ \times }}\;\frac{{{\rm{1}}\;{\rm{mol}}\;{\rm{MgB}}{{\rm{r}}_{\rm{2}}}}}{{{\rm{2}}\;\cancel{{{\rm{mol}}\;{\rm{AgN}}{{\rm{O}}_{\rm{3}}}}}}}\;{\rm{ = }}\;{\rm{0}}{\rm{.24}}\;{\rm{mol}}\]
And finally, convert the moles of MgBr2 to the volume of its solution using the molarity formula:
\[{\rm{V}}\left( {{\rm{MgB}}{{\rm{r}}_{\rm{2}}}} \right)\;{\rm{ = }}\;\frac{{\rm{n}}}{{\rm{M}}}\;{\rm{ = }}\;\frac{{{\rm{0}}{\rm{.24}}\;{\rm{mol}}}}{{{\rm{1}}{\rm{.80}}\;{\rm{mol/L}}}}{\rm{ = }}\;{\rm{0}}{\rm{.13}}\;{\rm{L}}\]
To convert this to mL, we multiply it by 1000, so it is 130 mL.
Check Also
1.
How many ml of 1.80 M MgBr2 is necessary to precipitate all the silver ions in 600.0 mL of 0.80 M AgNO3 solution?
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2.
How many ml of 2.70 M K2SO4 is required to precipitate all the barium ions in 500. mL of 0.650 M BaCl2 solution?
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3.
27.0 mL of a NaOH solution was needed to neutralize 0.326 g of KHP (KHC8H4O4). What is the molarity of the NaOH solution?
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4.
What volume of 0.150 M calcium hydroxide is required to neutralize 250.0 mL of 0.0350 M sulfuric acid?
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5.
How many mL of 0.250 M HCl will neutralize 60.00 mL of 0.060 M Ba(OH)2?
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6.
What is the mass in grams of AgCl precipitate if 20.0 mL of 0.150 M AgNO3 is added to 45.0 mL of 0.250 M MgCl2?
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7.
How many mL of 0.600 M K2SO4 was added to 400.0 mL of 0.500 M BaCl2 solution if 31.2 g of precipitate was collected?
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8.
When a 42.0-mL sample of a 0.85 M sodium chloride solution is mixed with 17.0 mL of a 0.650 M lead(II) nitrate solution, 2.45 g of a precipitate is formed:
2NaCl(aq) + Pb(NO3)2(aq) → PbCl2(s) + 2NaNO3(aq)
What are the theoretical and the percent yields of the reaction?
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9.
A 4.598-g sample of acetylsalicylic acid (the active ingredient in aspirin) was dissolved in water. It took 35.6 mL of a 0.717 M NaOH solution to neutralize the acid. Determine the molar mass of the acid considering that it is a monoprotic acid.
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10.
A student prepared a solution by dissolving 2.40-g mixture of calcium nitrate and calcium chloride. To this, a solution of silver nitrate was added dropwise until the mass of the precipitate stayed constant at 0.584 g. Determine the mass percent of calcium nitrate in the mixture.
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11.
Determine the products and calculate the mass of the precipitate that is formed after 177.2 mL of a 1.25 M Na2CrO4(aq) solution is mixed with 250 mL of a 1.25 M AgNO3(aq) solution.
AgNO3(aq) + Na2CrO4(aq) →
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