## General Chemistry

This is a set of practice problems in to help master the concept of limiting reactant which is critical in calculating the amount of product that can be obtained in a chemical reaction.

Remember, if the reactants are not in stoichiometric ratio, one of them is the limiting reactant (LR), and the other is in excess.

When solving problems involving a limiting reactant, keep in mind that we cannot look at the moles of the reactants and determine which one is the limiting reactant. The limiting reactant is the one that produces less product and not necessarily the one that is present in fewer amounts.

The correct approach is to calculate the moles of both reactants and, using molar conversions, determine which one produces less product.

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#### Practice

1.

Which statement about limiting reactant is correct?

a) The limiting reactant is the one in a smaller quantity.

b) The limiting reactant is the one in greater quantity.

c) The limiting reactant is the one producing less product.

d) The limiting reactant is the one producing more product.

The limiting reactant is the one producing less product than any of the other reactants.

2.

Find the limiting reactant for each initial amount of reactants.

4NH3 + 5O2 → 4NO + 6H2O

a) 2 mol of NH3 and 2 mol of O2

b) 2 mol of NHand 3 mol of O2

c) 3 mol of NH3 and 3 mol of O2

d) 3 mol of NH3 and 2 mol of O2

Note: This is not a multiple-choice question. Each row represents a separate question where you need to determine the limiting reactant.

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

Solution

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.

3.

How many g of hydrogen are leftover in producing ammonia when 14.0 g of nitrogen is reacted with 8.0 g of hydrogen?

N2(g) + 3 H2(g) → 2 NH3(g)

5 g

Solution

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

4.

When 140.0 g of AgNO3 was added to an aqueous solution of NaCl, 86.0 g of AgCl was collected as a white precipitate. Which salt was the limiting reactant in this reaction? How many grams of NaCl were present in the solution?

AgNO3(aq) + NaCl(aq) → AgCl(s) + NaNO3(aq)

NaCl was the limiting reactant and 35.0 g of it was present in the solution when AgNOwas added.

Solution

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 AgNO3.

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:

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

5.

How many grams of precipitate will be formed when 37.3 g KCl is reacted with 132.5 g Pb(NO3)2 according to the following chemical equation?

2 KCl(aq) + Pb(NO3)2(aq) → PbCl2(s) + 2 KNO3(aq)

69.5 g PbCl

Solution

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:

6.

How many grams of PClwill be produced if 38.5 g Clis reacted with 56.4 g P4 according to the following equation?

6Cl2(g) + P4(s) → 4PCl3(l)

169 g

Solution

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:

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

7.

How many grams of sulfur can be obtained if 12.6 g H2S is reacted with 14.6 g SO2 according to the following equation?

2H2S(g) + SO2(g) → 3S(s) + 2H2O(g)

17.8 g

Solution

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:

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

8.

The following equation represents the combustion of octane, C8H18, a component of gasoline:

2C8H18(g) + 25O2(g) → 16CO2(g) + 18H2O(g)

Will 356 g of oxygen be enough for the complete combustion of 954 g of octane?

356 g of oxygen is not enough for the complete combustion of 954 g of octane.

Solution

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.

9.

Consider the reaction between MnO2 and HCl:

MnO2 + 4HCl → MnCl2 + Cl2 + 2H2O

What is the theoretical yield of MnCl2 in grams when 165 g of MnO2 is added to a solution containing 94.2 g of HCl?

0.645 mol or 81.1 g MnCl2

Solution

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 MnCl2:

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