24.Limiting Reagent problems, including Stoichiometry, Actual yield, Theoretical

Question

24.Limiting Reagent problems, including Stoichiometry, Actual yield, Theoretical yield and Percent yield.

A. I react 5.0 grams of propane, C3H8, with excess oxygen to produce water and carbon dioxide. Produce the balanced combustion equation of propane and determine the theoretical yield of water. If only 6.8 grams of water are produced what is the percent yield of water?

B. A 2.00 g sample of ammonia is mixed with 4.00 g of oxygen. Which is the limiting reagent and how much excess reactant remains on completion of the reaction? What is the theoretical yield of NO. The equation below represents this reaction. ____ NH3(g) + ____ O2(g) ® ____ NO(g) + ____ H2O(g)

C. If 23.54 grams of nitrogen is reacted with 11.14 grams of hydrogen. Balance the following reaction. What is the theoretical yield of this reaction and which component is the limiting reagent? If only 8.26 grams of ammonia are produced what is the percent yield? ____ N2(g) + ____ H2(g) ® ____ NH3(g)

Explanation / Answer

Ans. A. Moles of propane = mass / molar mass = 5.0 g/ (44.0 g mol-1) = 0.1136 mol

Balanced reaction: C3H8 + 5 O2 ---------------> 3 CO2 + 4 H2O

Stoichiometry: 1 mol C3H8 combines with 5 mol O2 to form 3 mol CO2 and 4 mol H2O.

According to the stoichiometry of balanced reaction, 1 mol propane produced 4 mol water.

So,

Theoretical number of moles of water produced = 4 x moles of propane

= 4 x 0.1136 mol = 0.4545 mol

Theoretical mass of water produced = theoretical mol of water x molar mass

                                    = 0.4545 mol x (18.0 g mol-1) = 8.18 g

% yield = (Actual mass of water obtained / theoretical mass of water obtained) x 100

            = (6.8 g / 8.18 g) x 100

            = 83.13 %

Ans B. Moles of NH3 = mass / molar mass = 2.0 g/ (17.0 g mol-1) = 0.1176 mol

            Moles of O2 = 4.0 g / (32.0 g mol-1) = 0.125 mol

Balanced reaction: 4 NH3 + 5 O2 ---------------> 4 NO + 6 H2O

Stoichiometry: 4 mol NH3 combines with 5 mol O2 to form 4 mol NO and 6 mol H2O.

Theoretical molar ration: moles of NH3 : O2 = 4 : 5 = 1: 1.25

Experimental (given) ratio : NH3 : O2 = 0.1176 : 0.125 Convert this ration in such a way that at least one value become equal to that of any one value of theoretical value. For example, multiply experimental ratio by a factor “10”. Note that dividing or multiplying does not affect the ration.

            So, corrected ration, NH3 : O2 = (0.1176 : 0.125) x 10= 1.176 : 1.25

Now, compare the theoretical ratio with corrected experimental ratio. Note that NH3 is in excess because it has 1.176 moles in the reaction mixture compared to 1.0 mole required in theoretical stoichiometry.

Therefore,

            Limiting reagent = O2

            Reagent in excess = NH3

As the product formation follows limiting reagent stoichiometry, the number of moles of NO produced is proportional to moles of NH3.

According to the stoichiometry of balanced reaction, 5 mol O2 produced 4 mol NO.

So, moles of NO produced = (4/ 5) x 0.125 mol = 0.1 mol

Mass of NO produced = Moles x molar mass = 0.1 mol x (30.0 g mol-1) = 3.0 g

Thus, theoretical yield of NO = 3.0 g

Ans. C. Moles of N2 = mass / molar mass = 23.54 g/ (28.0 g mol-1) = 0.84 mol

            Moles of H2 = 11.14 g / (2.0 g mol-1) = 5.57 mol

Balanced reaction: N2 + 3 H2 ---------------> 2 NH3

Stoichiometry: 1 mol N2 combines with 3 mol H2 to form 2 mol NH3.

Theoretical molar ration: moles of N2 : H2 = 1: 3

Experimental (given) ratio : N2 : H2 = 0.84 : 5.

            So, corrected ration, N2 : H2 = (0.84 : 5.57) / 0.84 = 1 : 6.63

Now, compare the theoretical ratio with corrected experimental ratio. Note that H2 is in excess because it has 6.63 moles in the reaction mixture compared to 3.0 mole required in theoretical stoichiometry.

Therefore,

            Limiting reagent = N2

            Reagent in excess = H2

As the product formation follows limiting reagent stoichiometry, the number of moles of NH3 produced is proportional to moles of N2.

According to the stoichiometry of balanced reaction, 1 mol N2 produced 2 mol NH3.

So, moles of NH3 produced = (2) x 0.84 mol = 1.68 mol

Mass of NH3 produced = Moles x molar mass = 1.68 mol x (17.0 g mol-1) = 28.56 g

Thus, theoretical yield of NO = 28.56 g

Given, Actual yield = 8.26 g

% yield = (Actual yield/ theoretical yield) x 100

            = (8.26 g / 28.56 g) x 100

            = 28.92 %

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