The limiting reactant is the one producing less product. So, to determine the limiting reactant for each pair, we need to pick a product (NO or H₂O) and see whether NH₃ or O₂ will produce less product for the given mole ratio.

Let’s do the calculations based on the amount of NO that can be formed.

a)

\[{\rm{n}}\;\left( {{\rm{NO}}} \right)\;{\rm{ = }}\;{\rm{2 }}\cancel{{{\rm{mol}}\;{\rm{N}}{{\rm{H}}_{\rm{3}}}}}\;{\rm{ \times }}\;\frac{{{\rm{4}}\;{\rm{mol}}\;{\rm{NO}}}}{{{\rm{4}}\;\cancel{{{\rm{mol}}\;{\rm{N}}{{\rm{H}}_{\rm{3}}}}}}}\;{\rm{ = }}\;{\rm{2}}\;{\rm{mol}}\]

\[{\rm{n}}\;\left( {{\rm{NO}}} \right)\;{\rm{ = }}\;{\rm{2 }}\cancel{{{\rm{mol}}\;{{\rm{O}}_2}}}\;{\rm{ \times }}\;\frac{{{\rm{4}}\;{\rm{mol}}\;{\rm{NO}}}}{{{\rm{5}}\;\cancel{{{\rm{mol}}\;{{\rm{O}}_2}}}}}\;{\rm{ = }}\;{\rm{1}}{\rm{.6}}\;{\rm{mol}}\]

So, for (a), O₂ is the limiting reactant.

b)

\[{\rm{n}}\;\left( {{\rm{NO}}} \right)\;{\rm{ = }}\;{\rm{2 }}\cancel{{{\rm{mol}}\;{\rm{N}}{{\rm{H}}_{\rm{3}}}}}\;{\rm{ \times }}\;\frac{{{\rm{4}}\;{\rm{mol}}\;{\rm{NO}}}}{{{\rm{4}}\;\cancel{{{\rm{mol}}\;{\rm{N}}{{\rm{H}}_{\rm{3}}}}}}}\;{\rm{ = }}\;{\rm{2}}\;{\rm{mol}}\]

\[{\rm{n}}\;\left( {{\rm{NO}}} \right)\;{\rm{ = }}\;{\rm{3 }}\cancel{{{\rm{mol}}\;{{\rm{O}}_2}}}\;{\rm{ \times }}\;\frac{{{\rm{4}}\;{\rm{mol}}\;{\rm{NO}}}}{{{\rm{5}}\;\cancel{{{\rm{mol}}\;{{\rm{O}}_2}}}}}\;{\rm{ = }}\;{\rm{2}}{\rm{.4}}\;{\rm{mol}}\]

So, for (b), NO is the limiting reactant.

c)

\[{\rm{n}}\;\left( {{\rm{NO}}} \right)\;{\rm{ = }}\;{\rm{3 }}\cancel{{{\rm{mol}}\;{\rm{N}}{{\rm{H}}_{\rm{3}}}}}\;{\rm{ \times }}\;\frac{{{\rm{4}}\;{\rm{mol}}\;{\rm{NO}}}}{{{\rm{4}}\;\cancel{{{\rm{mol}}\;{\rm{N}}{{\rm{H}}_{\rm{3}}}}}}}\;{\rm{ = }}\;{\rm{3}}\;{\rm{mol}}\]

\[{\rm{n}}\;\left( {{\rm{NO}}} \right)\;{\rm{ = }}\;{\rm{3 }}\cancel{{{\rm{mol}}\;{{\rm{O}}_2}}}\;{\rm{ \times }}\;\frac{{{\rm{4}}\;{\rm{mol}}\;{\rm{NO}}}}{{{\rm{5}}\;\cancel{{{\rm{mol}}\;{{\rm{O}}_2}}}}}\;{\rm{ = }}\;{\rm{2}}{\rm{.4}}\;{\rm{mol}}\]

So, for (c), O₂ is the limiting reactant.

d)

\[{\rm{n}}\;\left( {{\rm{NO}}} \right)\;{\rm{ = }}\;{\rm{3 }}\cancel{{{\rm{mol}}\;{\rm{N}}{{\rm{H}}_{\rm{3}}}}}\;{\rm{ \times }}\;\frac{{{\rm{4}}\;{\rm{mol}}\;{\rm{NO}}}}{{{\rm{4}}\;\cancel{{{\rm{mol}}\;{\rm{N}}{{\rm{H}}_{\rm{3}}}}}}}\;{\rm{ = }}\;{\rm{3}}\;{\rm{mol}}\]

\[{\rm{n}}\;\left( {{\rm{NO}}} \right)\;{\rm{ = }}\;{\rm{2 }}\cancel{{{\rm{mol}}\;{{\rm{O}}_2}}}\;{\rm{ \times }}\;\frac{{{\rm{4}}\;{\rm{mol}}\;{\rm{NO}}}}{{{\rm{5}}\;\cancel{{{\rm{mol}}\;{{\rm{O}}_2}}}}}\;{\rm{ = }}\;{\rm{1}}{\rm{.6}}\;{\rm{mol}}\]

So, for (d), O₂ is the limiting reactant.

The shorter way of determining the limiting reactant is to divide the moles of a given reactant by its coefficient in the chemical reaction. This is the effective quantity of the reactant as it takes into consideration not only the amount of the reactant but also its molar ratio in the reaction.

Let’s put the equation and the moles here and determine the LR by the ratio of the moles and coefficient:

4NH₃ + 5O₂ → 4NO + 6H₂O

a) 2 mol of NH₃ and 2 mol of O₂

b) 2 mol of NH₃ and 3 mol of O₂

c) 3 mol of NH₃ and 3 mol of O₂

d) 3 mol of NH₃ and 2 mol of O₂

So, for (a), we are comparing 2/4 mol of NH₃ with 2/5 mol of O₂, and therefore, O₂ is the LR.

For (b), we are comparing 2/4 mol of NH₃ with 3/5 mol of O₂, and therefore, NH₃ is the LR.

For (c), we are comparing 3/4 mol of NH₃ with 3/5 mol of O₂, and therefore, O₂ is the LR.

For (d),  3 mol of NH₃ and 3/4 mol of NH₃ with 2/5 mol of O₂, and therefore, O₂ is the LR.

As stated in the problem, there is going to be some H₂ left over after the reaction is complete, so this tells us that H₂ is in excess and N₂ is the limiting reactant. Remember, limiting reactant is consumed completely in a chemical reaction.

Remember also that stoichiometric calculations need to be done based on the moles of limiting reactant, so let’s first determine the limiting reactant.

Limiting reactant: Now, let’s determine which reactant will produce less ammonia. It would be very beneficial to practice and determine the moles in the chemical reaction without doing mole ratio calculations.

a) Alternative to dimensional analysis. For example, the chemical equation indicates that from every one mole of N₂ there is 2 moles of NH₃ forming. This means that every 2 moles of N₂ will produce 4 moles of NH₃, every 3 moles of N₂ will produce 6 moles of NH₃, every 10 moles of N₂ will produce 20 moles of NH₃ etc. Ammonia is always two times more.

So, in this case, 0.50 mol N₂ will produce 1.0 mol NH₃.

On the other hand, every 3 moles of H₂ produces 2 moles of NH₃, so ammonia is always going to be 1.5 times less than the H₂.

We can put the moles on top of H₂ and mark the moles of NH₃ as X.

               4.0 mol       X mol
N₂(g) + 3 H₂(g) → 2 NH₃(g)

Then cross multiply to solve for the X.

We can put the results together into one equation:

0.50 mol                   1.0 mol
               4.0 mol      2.7mol
N2(g) + 3 H2(g) → 2 NH3(g)

Based on the moles of NH₃, N₂ is the limiting reactant in this reaction.

b) Dimensional analysis.
The method shown above is a quicker way of doing stoichiometry calculations but of course you can also use the dimensional analysis approach:

Based on N₂, there will be:

Based on H₂, there will be:

As expected, both approaches lead to the same answer and N₂ is the limiting reactant. We now need to calculate how much H₂ is left over after the reaction is complete and in order to do it, we will first calculate how much of it is consumed in the reaction based on the moles of the limiting reactant.

This is how much H₂ reacted when the limiting reactant (N₂) was consumed completely. To find how much is left, we need to subtract the initial and consumed amounts:

n (H₂) = 4 mol (initial) – 1.5 mol (reacted) = 2.5 mol H₂ left over

mass H₂ = 2.5 mol x 2.0 g/mol = 5.0 g

We need to find the moles of AgCl which will allow us to calculate how much AgNO₃ and NaCl had reacted in the reaction.

\[{\rm{n}}\;{\rm{(AgCl)}}\;{\rm{ = }}\;{\rm{86}}{\rm{.0}}\;{\rm{g}}\;{\rm{ \times }}\;\frac{{{\rm{1}}\;{\rm{mol}}}}{{{\rm{143}}{\rm{.3}}\;{\rm{g}}}}\;{\rm{ = }}\;{\rm{0}}{\rm{.600}}\;{\rm{mol}}\]

The stoichiometric ratio of all the reactants and products is1:1, and therefore, there were 0.600 mol of AgNO₃ and 0.600 mol of NaCl participating in the reaction.

Now, let’s calculate the moles of 140.0 g of AgNO₃ and see whether it corresponds to 0.600 mol. If it is more than that, then AgNO₃ was added in excess and NaCl must be the limiting reactant:

\[{\rm{n}}\;{\rm{(AgN}}{{\rm{O}}_{\rm{3}}}{\rm{)}}\;{\rm{ = }}\;{\rm{140}}{\rm{.0}}\;{\rm{g}}\;{\rm{ \times }}\;\frac{{{\rm{1}}\;{\rm{mol}}}}{{{\rm{169}}{\rm{.9}}\;{\rm{g}}}}\;{\rm{ = }}\;{\rm{0}}{\rm{.824}}\;{\rm{mol}}\]

So, 0.824 mol of AgNO₃ was added to the solution, but only 0.600 mol had reacted which means that is how much NaCl was present in the solution, and it was not enough to react with all the AgNO_3.

And this confirms that AgNO₃ was added in excess and NaCl is the limiting reactant.

In the last step, we determine the mass of 0.600 mol NaCl that was in the solution when AgNO₃ was added:

\[{\rm{m}}\;{\rm{(NaCl)}}\;{\rm{ = }}\;{\rm{0}}{\rm{.600}}\;{\rm{mol}}\;{\rm{ \times }}\;\frac{{{\rm{58}}{\rm{.4}}\;{\rm{g}}}}{{{\rm{1}}\;{\rm{mol}}}}\;{\rm{ = }}\;{\rm{35}}{\rm{.0}}\;{\rm{g}}\]

In order to find the amount of products formed or reactants consumed in any chemical equation, we need to first calculate the moles from the corresponding masses. There are exceptions where the calculations can be done based on the volumes but we will not discuss them here.

Moles can be calculated from the following formula:

Second, we need to identify the precipitate and a solubility table will tell us that PbCl₂ is the precipitate which is also indicated by the (s – solid) notation in the chemical equation.

At this point we have the moles of both reactants, but in order to determine how much product is formed, we need to always use the limiting reactant.

Let’s determine independently how much PbCl₂ could be formed from each reactant.

Based on KCl:

Based on Pb(NO₃)₂:

The limiting reactant in this reaction is KCl and based on the mole ratio, 0.25 mole of PbCl2 will be formed.

The last step is to convert the moles to grams:

From the formula

we can use cross multiplication and find that the mass is equal to:

m = n x M

so, m (PbCl₂) = 0.250 mol x 278.1 g/mol = 69.5 g PbCl₂

Alternatively, you can use dimensional analysis to calculate the mass:

When the amounts of both reactants are given, you must determine the LR and then, based on the moles of the LR, determine the amount of product(s) that can be formed.

The moles of Cl₂ and P₄ are:

\[{\rm{n}}\;{\rm{(C}}{{\rm{l}}_{\rm{2}}}{\rm{)}}\;{\rm{ = }}\;{\rm{38}}{\rm{.5}}\;{\rm{g}}\;{\rm{ \times }}\;\frac{{{\rm{1}}\;{\rm{mol}}}}{{{\rm{70}}{\rm{.9}}\;{\rm{g}}}}\;{\rm{ = }}\;{\rm{1}}{\rm{.84}}\;{\rm{mol}}\]

\[{\rm{n}}\;{\rm{(}}{{\rm{P}}_{\rm{4}}}{\rm{)}}\;{\rm{ = }}\;{\rm{56}}{\rm{.4}}\;{\rm{g}}\;{\rm{ \times }}\;\frac{{{\rm{1}}\;{\rm{mol}}}}{{{\rm{123}}{\rm{.9}}\;{\rm{g}}}}\;{\rm{ = }}\;{\rm{0}}{\rm{.455}}\;{\rm{mol}}\]

Next, we calculate how much PCl₃ can be formed from each reactant:

\[{\rm{n}}\;\left( {{\rm{PC}}{{\rm{l}}_{\rm{3}}}} \right)\;{\rm{ = }}\;{\rm{1}}{\rm{.84 }}\cancel{{{\rm{mol}}\;{\rm{C}}{{\rm{l}}_{\rm{2}}}}}\;{\rm{ \times }}\;\frac{{{\rm{4}}\;{\rm{mol}}\;{\rm{PC}}{{\rm{l}}_{\rm{3}}}}}{{{\rm{6}}\;\cancel{{{\rm{mol}}\;{\rm{C}}{{\rm{l}}_{\rm{2}}}}}}}\;{\rm{ = }}\;{\rm{1}}{\rm{.23}}\;{\rm{mol}}\]

\[{\rm{n}}\;\left( {{\rm{PC}}{{\rm{l}}_{\rm{3}}}} \right)\;{\rm{ = }}\;{\rm{0}}{\rm{.455 }}\cancel{{{\rm{mol}}\;{{\rm{P}}_{\rm{4}}}}}\;{\rm{ \times }}\;\frac{{{\rm{4}}\;{\rm{mol}}\;{\rm{PC}}{{\rm{l}}_{\rm{3}}}}}{{{\rm{1}}\;\cancel{{{\rm{mol}}\;{{\rm{P}}_{\rm{4}}}}}}}\;{\rm{ = }}\;{\rm{1}}{\rm{.82}}\;{\rm{mol}}\]

Cl₂ gives less PCl₃ and therefore, it is the LR and 1.23 mol of PCl₃ can be produced in this reaction.

The mass of PCl₃ is:

\[{\rm{m}}\;{\rm{(PC}}{{\rm{l}}_{\rm{3}}}{\rm{)}}\;{\rm{ = }}\;{\rm{1}}{\rm{.23}}\;{\rm{mol}}\;{\rm{ \times }}\;\frac{{{\rm{137}}{\rm{.3}}\;{\rm{g}}}}{{{\rm{1}}\;{\rm{mol}}}}\;{\rm{ = }}\;{\rm{169}}\;{\rm{g}}\]

When the amount of both reactants is given, you must determine the LR and then, based on the moles of the LR, determine the amount of product(s) that can be formed.

The moles of H₂S and SO₂ are:

\[{\rm{n}}\;{\rm{(}}{{\rm{H}}_{\rm{2}}}{\rm{S)}}\;{\rm{ = }}\;{\rm{12}}{\rm{.6}}\;{\rm{g}}\;{\rm{ \times }}\;\frac{{{\rm{1}}\;{\rm{mol}}}}{{{\rm{34}}{\rm{.1}}\;{\rm{g}}}}\;{\rm{ = }}\;{\rm{0}}{\rm{.370}}\;{\rm{mol}}\]

\[{\rm{n}}\;{\rm{(S}}{{\rm{O}}_{\rm{2}}}{\rm{)}}\;{\rm{ = }}\;{\rm{14}}{\rm{.6}}\;{\rm{g}}\;{\rm{ \times }}\;\frac{{{\rm{1}}\;{\rm{mol}}}}{{{\rm{64}}{\rm{.1}}\;{\rm{g}}}}\;{\rm{ = }}\;{\rm{0}}{\rm{.228}}\;{\rm{mol}}\]

Next, we calculate how much PCl₃ can be formed from each reactant:

\[{\rm{n}}\;\left( {\rm{S}} \right)\;{\rm{ = }}\;{\rm{0}}{\rm{.370 }}\cancel{{{\rm{mol}}\;{{\rm{H}}_{\rm{2}}}{\rm{S}}}}\;{\rm{ \times }}\;\frac{{{\rm{3}}\;{\rm{mol}}\;{\rm{S}}}}{{{\rm{2}}\;\cancel{{{\rm{mol}}\;{{\rm{H}}_{\rm{2}}}{\rm{S}}}}}}\;{\rm{ = }}\;{\rm{0}}{\rm{.555}}\;{\rm{mol}}\]

\[{\rm{n}}\;\left( {\rm{S}} \right)\;{\rm{ = }}\;{\rm{0}}{\rm{.228 }}\cancel{{{\rm{mol}}\;{\rm{S}}{{\rm{O}}_{\rm{2}}}}}\;{\rm{ \times }}\;\frac{{{\rm{3}}\;{\rm{mol}}\;{\rm{S}}}}{{{\rm{1}}\;\cancel{{{\rm{mol}}\;{\rm{S}}{{\rm{O}}_{\rm{2}}}}}}}\;{\rm{ = }}\;{\rm{0}}{\rm{.684}}\;{\rm{mol}}\]

H₂S gives less S and therefore, it is the LR and 0.555 mol of S can be produced in this reaction.

The mass of S is:

\[{\rm{m}}\;{\rm{(S)}}\;{\rm{ = }}\;{\rm{0}}{\rm{.555}}\;{\rm{mol}}\;{\rm{ \times }}\;\frac{{{\rm{32}}{\rm{.1}}\;{\rm{g}}}}{{{\rm{1}}\;{\rm{mol}}}}\;{\rm{ = }}\;{\rm{17}}{\rm{.8}}\;{\rm{g}}\]

First, calculate the moles of the reactants:

\[{\rm{n}}\;{\rm{(}}{{\rm{O}}_{\rm{2}}}{\rm{)}}\;{\rm{ = }}\;{\rm{356}}\;{\rm{g}}\;{\rm{ \times }}\;\frac{{{\rm{1}}\;{\rm{mol}}}}{{{\rm{32}}{\rm{.0}}\;{\rm{g}}}}\;{\rm{ = }}\;{\rm{11}}{\rm{.4}}\;{\rm{mol}}\]

\[{\rm{n}}\;{\rm{(}}{{\rm{C}}_{\rm{8}}}{{\rm{H}}_{{\rm{18}}}}{\rm{)}}\;{\rm{ = }}\;{\rm{954}}\;{\rm{g}}\;{\rm{ \times }}\;\frac{{{\rm{1}}\;{\rm{mol}}}}{{{\rm{114}}{\rm{.2}}\;{\rm{g}}}}\;{\rm{ = }}\;{\rm{8}}{\rm{.35}}\;{\rm{mol}}\]

And now let’s see how many moles of O₂ are needed to react with 8.35 mol of C₈H₁₈:

\[{\rm{n}}\;\left( {{{\rm{O}}_{\rm{2}}}} \right)\;{\rm{ = }}\;{\rm{8}}{\rm{.35 }}\cancel{{{\rm{mol}}\;{{\rm{C}}_{\rm{8}}}{{\rm{H}}_{{\rm{18}}}}}}\;{\rm{ \times }}\;\frac{{{\rm{25}}\;{\rm{mol}}\;{{\rm{O}}_{\rm{2}}}}}{{{\rm{2}}\;\cancel{{{\rm{mol}}\;{{\rm{C}}_{\rm{8}}}{{\rm{H}}_{{\rm{18}}}}}}}}\;{\rm{ = }}\;{\rm{104}}\;{\rm{mol}}\]

104 moles of O₂ are needed to react with only 8.35 mol of C₈H₁₈ and this is because of the 25:2 mole ratio. So, 356 g of oxygen is not enough for the complete combustion of 954 g of octane.

The theoretical yield is the amount of product that can be formed, so this problem is similar to the ones we have been working on.

The amounts of both reactants are given, so we determine the LR and then, based on the moles of the LR, determine the amount of product(s) that can be formed.

The moles of MnO₂ and HCl are:

\[{\rm{n}}\;{\rm{(Mn}}{{\rm{O}}_{\rm{2}}}{\rm{)}}\;{\rm{ = }}\;{\rm{165}}\;{\rm{g}}\;{\rm{ \times }}\;\frac{{{\rm{1}}\;{\rm{mol}}}}{{{\rm{86}}{\rm{.9}}\;{\rm{g}}}}\;{\rm{ = }}\;{\rm{1}}{\rm{.90}}\;{\rm{mol}}\]

\[{\rm{n}}\;{\rm{(HCl)}}\;{\rm{ = }}\;{\rm{94}}{\rm{.2}}\;{\rm{g}}\;{\rm{ \times }}\;\frac{{{\rm{1}}\;{\rm{mol}}}}{{{\rm{36}}{\rm{.5}}\;{\rm{g}}}}\;{\rm{ = }}\;{\rm{2}}{\rm{.58}}\;{\rm{mol}}\]

Next, we calculate how much MnCl₂ can be formed from each reactant:

\[{\rm{n}}\;\left( {{\rm{MnC}}{{\rm{l}}_{\rm{2}}}} \right)\;{\rm{ = }}\;{\rm{1}}{\rm{.90 }}\cancel{{{\rm{mol}}\;{\rm{Mn}}{{\rm{O}}_{\rm{2}}}}}\;{\rm{ \times }}\;\frac{{{\rm{1}}\;{\rm{mol}}\;{\rm{MnC}}{{\rm{l}}_{\rm{2}}}}}{{{\rm{1}}\;\cancel{{{\rm{mol}}\;{\rm{Mn}}{{\rm{O}}_{\rm{2}}}}}}}\;{\rm{ = }}\;{\rm{1}}{\rm{.90}}\;{\rm{mol}}\]

\[{\rm{n}}\;\left( {{\rm{MnC}}{{\rm{l}}_{\rm{2}}}} \right)\;{\rm{ = }}\;{\rm{2}}{\rm{.58 }}\cancel{{{\rm{mol}}\;{\rm{HCl}}}}\;{\rm{ \times }}\;\frac{{{\rm{1}}\;{\rm{mol}}\;{\rm{MnC}}{{\rm{l}}_{\rm{2}}}}}{{{\rm{4}}\;\cancel{{{\rm{mol}}\;{\rm{HCl}}}}}}\;{\rm{ = }}\;{\rm{0}}{\rm{.645}}\;{\rm{mol}}\]

HCl gives less product so, it is the LR, and therefore, 0.645 mol MnCl₂ is the theoretical yield of the reaction.

If you are asked to find the theoretical yield in grams, then convert the moles to grams using the molar mass of MnCl_2:

\[{\rm{m}}\;{\rm{(MnC}}{{\rm{l}}_{\rm{2}}}{\rm{)}}\;{\rm{ = }}\;{\rm{0}}{\rm{.645}}\;{\rm{mol}}\;{\rm{ \times }}\;\frac{{{\rm{125}}{\rm{.8}}\;{\rm{g}}}}{{{\rm{1}}\;{\rm{mol}}}}\;{\rm{ = }}\;{\rm{81}}{\rm{.1}}\;{\rm{g}}\]