### StoichiometryLimiting Reagent Problems solved using only dimensional analysis

Example #1: Given the following reaction: 2Mg + O2 ---> 2MgO

What is the limiting reactant if 2.20 g of Mg reacts with 4.50 L of oxygen at STP?

Solution:

1) From the point of view of Mg reacting with O2:

 2.20 g Mg 1 mol Mg 1 mol O2 22.414 L O2 –––––––– x –––––––––– x –––––––– x ––––––––– = 1.01 L O2 1 24.305 g Mg 2 mol Mg 1 mol O2

All 2.20 g of Mg is consumed by 1.01 L of O2. Mg is shown to be the limiting reagent.

2) From the point of view of O2 reacting with Mg:

 4.50 L O2 1 mol O2 2 mol Mg 24.305 g Mg –––––––– x –––––––––– x –––––––– x ––––––––––– = 9.76 g Mg 1 22.414 L O2 1 mol O2 1 mol Mg

Consuming all 4.50 L of O2 would require 9.76 g of Mg for complete reaction. Mg is shown to be the limiting reagent.

3) From the point of view of Mg producing MgO:

 2.20 g Mg 1 mol Mg 2 mol MgO –––––––– x –––––––––– x ––––––––– = 0.0905 mol MgO 1 24.305 g Mg 2 mol Mg

Consuming all 2.20 g of Mg produces 0.0905 mol of MgO.

4) From the point of view of O2 producing MgO:

 4.50 L O2 1 mol O2 2 mol MgO –––––––– x –––––––––– x ––––––––– = 0.402 mol MgO 1 22.414 L O2 1 mol O2

Consuming all 4.50 L of O2 produces 0.402 mol of MgO.

5) A comparison of #3 and #4 shows that Mg is the limiting reagent.

6) Note that either calculation of #1 or #2 is sufficient to show Mg is the limiting reagent. Whereas, both #3 and #4 are required to show (by comparison of results) that Mg is the limiting reagent.

7) Division of each reactant's moles by its coefficient leads to identifying the limiting reagent:

 2.20 g Mg 1 mol Mg 1 –––––––– x –––––––––– x –––––––– = 0.0905 <--- the smaller result identifies the limiting reageant 1 24.305 g Mg 2 mol Mg

 4.50 L O2 1 mol O2 1 –––––––– x –––––––––– x –––––––– = 0.201 1 22.414 L O2 1 mol O2

Both calculations must be done with the lowest value being the limiting reagent.

Example #2: Given the following reaction:

CH4 + 2H2O ---> CO2 + 4H2
(a) How many liters of hydrogen can be produced from the reaction between 80.0 g of CH4 and 16.3 g of water? (b) Determine the limiting reagent.

Solution to (b):

1) Determine how much one of the reactants needs of the other:

 80.0 g CH4 1 mol CH4 2 mol H2O 18.015 g H2O –––––––– x –––––––––– x –––––––– x ––––––––––– = 180. g H2O 1 16.033 g CH4 1 mol CH4 1 mol H2O

 16.3 g H2O 1 mol H2O 1 mol CH4 16.033 g CH4 –––––––– x –––––––––– x –––––––– x ––––––––––– = 7.25 g CH4 1 18.015 g H2O 2 mol H2O 1 mol CH4

2) You can use either one of the above calculations to determine the limiting reagent. Note that:

in the first calculation, far more water is required than what is present. Water is limiting.

In the second calculation, less CH4 is used than what is present, meaning CH4 is in excess. This leads immediately to the conclusion that water is the limiting reagent.

3) You may also determine the limiting reagent by calculating how much product is produced by each reactant (assuming each one, in turn, is the limiting reagent):

 80.0 g CH4 1 mol CH4 4 mol H2 22.414 L H2 –––––––– x –––––––––– x –––––––– x ––––––––– = 447 L H2 1 16.033 g CH4 1 mol CH4 1 mol H2

 16.3 g H2O 1 mol H2O 4 mol H2 22.414 L H2 –––––––– x –––––––––– x –––––––– x ––––––––– = 40.5 L H2 1 18.015 g H2O 2 mol H2O 1 mol H2

The lesser of the two answers identifies the limiting reagent to be water.

4) Stopping the calculation at moles is sufficient to identify the limiting reagent:

 80.0 g CH4 1 mol CH4 4 mol H2 –––––––– x –––––––––– x –––––––– = 20.0 mol H2 1 16.033 g CH4 1 mol CH4

 16.3 g H2O 1 mol H2O 4 mol H2 –––––––– x –––––––––– x –––––––– = 1.81 mol H2 1 18.015 g H2O 2 mol H2O

The lesser of the two answers identifies the limiting reagent to be water.

Example #3: Given this reaction: 2NaCl + Pb(NO3)2 ---> 2NaNO3 + PbCl2

(a) How many grams of lead(II) chloride are produced from the reaction of 15.3 g of NaCl and 60.8 g of Pb(NO3)2? (b) What is the limiting reactant? (c) How much excess is left over?

Solution:

1) Let us determine answer (b), the limiting reagent, first.

Method one: determine how much one of the reactant needs of the other.

 15.3 g NaCl 1 mol NaCl 1 mol Pb(NO3)2 331.208 g Pb(NO3)2 –––––––––– x ––––––––––– x ––––––––––––– x –––––––––––––––– = 43.4 g Pb(NO3)2 <--- will use this to calculate answer to (c) 1 58.443 g NaCl 2 mol NaCl 1 mol Pb(NO3)2

 60.8 g Pb(NO3)2 1 mol Pb(NO3)2 2 mol NaCl 58.443 g NaCl ––––––––––––– x ––––––––––––––––– x –––––––––––– x –––––––––––– = 21.4 g NaCl 1 331.208 g Pb(NO3)2 1 mol Pb(NO3)2 1 mol NaCl

We do not have enough NaCl (21.4 g required to react with the full 60.8 g, but we have only 15.3 g. NaCl is the limiting reagent. Or, the 15.3 g of NaCl only react with 43.4 g of lead(II) nitrate, leaving it in excess. Therefore, NaCl is the limiting reagent.

Method two: how much product is produced by each reactant.

 15.3 g NaCl 1 mol NaCl 1 mol PbCl2 278.106 g PbCl2 –––––––––– x ––––––––––– x –––––––––– x –––––––––––––– = 36.4 g PbCl2 <--- this is the answer to (a) 1 58.443 g NaCl 2 mol NaCl 1 mol PbCl2

 60.8 g Pb(NO3)2 1 mol Pb(NO3)2 1 mol PbCl2 278.106 g PbCl2 ––––––––––––– x –––––––––––––––– x –––––––––––– x ––––––––––––– = 51.0 g PbCl2 1 331.208 g Pb(NO3)2 1 mol Pb(NO3)2 1 mol PbCl2

NaCl is the limiting reagent because it makes the least amount of product.

Method three: compare moles of product rather than grams of product.

 15.3 g NaCl 1 mol NaCl 1 mol PbCl2 –––––––––– x ––––––––––– x –––––––––––––– = 0.131 mol PbCl2 1 58.443 g NaCl 2 mol NaCl

 60.8 g Pb(NO3)2 1 mol Pb(NO3)2 1 mol PbCl2 ––––––––––––– x –––––––––––––––– x ––––––––––––– = 0.184 mol PbCl2 1 331.208 g Pb(NO3)2 1 mol Pb(NO3)2

NaCl is the limiting reagent because it makes the least amount of product.

2) The answer to (a) is found in the first calculation under 'method two' just above.

3) The anser to (c) is this:

60.8 g − 43.4 g = 17.4 g Pb(NO3)2

The amount of Pb(NO3)2 used up is found in the first calculation under 'method one' just above.

Example #4: 2.5 mol of CaO reacts with 3.0 mol of SO2 and 2.0 mol of O2 to produce CaSO4. The balanced equation:

2CaO + 2SO2 + O2 ---> 2CaSO4

Determine the limiting reagent.

Solution:

1) Let each reactant produce the product:

 2.5 mol CaO 2 mol CaSO4 –––––––––– x –––––––– = 2.5 mol CaSO4 1 2 mol CaO

 3.0 mol SO2 2 mol CaSO4 –––––––––– x –––––––– = 3 mol CaSO4 1 2 mol SO2

 2.0 mol O2 2 mol CaSO4 –––––––––– x –––––––– = 4 mol CaSO4 1 1 mol O2

2) CaO produces the least amount of product. The CaO would run out before the SO2 or the O2 run out.

CaO is the limiting reagent.

Example #5: Solid calcium carbonate, CaCO3, is able to remove sulphur dioxide from waste gases by this reaction (balanced as written):

CaCO3 + SO2 + other reactants ---> CaSO3 + other products

In a particular experiment, 255 g of CaCO3 was exposed to 135 g of SO2 in the presence of an excess amount of the other chemicals required for the reaction.

(a) What is the theoretical yield of CaSO3?

(b) If only 198 g of CaSO3 was isolated from the products, what was the percentage yield of CaSO3 in this experiment?

Solution:

1) Allow each given amount of reactant to produce the product. The smaller amount will be the theoretical yield.

 255 g CaCO3 1 mol CaCO3 1 mol CaSO3 120.14 g CaSO3 ––––––––––– x ––––––––––––– x ––––––––––– x ––––––––––––– = 306 g CaSO3) 1 100.086 g CaCO3 1 mol CaCO3 1 mol CaSO3

 135 g SO2 1 mol SO2 1 mol CaSO3 120.14 g CaSO3 ––––––––– x ––––––––––– x ––––––––––– x ––––––––––––– = 253 g CaSO3) 1 64.063 g SO2 1 mol SO2 1 mol CaSO3

2) Percent yield:

(198 g / 253 g) * 100 = 78.3%