Answer to Question #235476 in Chemistry for Dotun

Question

169 g FeCr2O4, 298 g K2CO3 and an excess of O2 (g) are sealed in a reaction vessel and allowed to react at high temperature. The amount of K2CrO4 obtained is 194 g. Calculate the percent yield of K2CrO4.

4 FeCr2O4  +  8 K2CO3  + 7 O2 → 8 K2CrO4 + 2 Fe2O3 + 8 CO2 

( Molar mass: FeCr2O4 = 223.84, K2CO3=138.21, K2CrO4 =194.19 g/mole )

Answer 1

Solution:

Calculate moles of each reactant:

(169 g FeCr₂O₄) × (1 mol FeCr₂O₄ / 223.84 g FeCr₂O₄) = 0.755 mol FeCr₂O₄

(298 g K₂CO₃) × (1 mol K₂CO₃ / 138.21 g K₂CO₃) = 2.156 mol K₂CO₃

Balanced chemical equation:

4FeCr₂O₄ + 8K₂CO₃ + 7O₂ → 8K₂CrO₄ + 2Fe₂O₃ + 8CO₂ 

According to stoichiometry:

4 mol of FeCr₂O₄ reacts with 8 mol of K₂CO₃

Thus, 0.755 mol FeCr₂O₄ reacts with:

(0.755 mol FeCr₂O₄) × (8 mol K₂CO₃ / 4 mol FeCr₂O₄) = 1.51 mol K₂CO₃

However, initially there is 2.156 mol of K₂CO₃ (according to the task).

Therefore, FeCr₂O₄ acts as limiting reagent and K₂CO₃ is excess reagent.

According to stoichiometry:

4 mol of FeCr₂O₄ produces 8 mol of K₂CrO₄

Thus, 0.755 mol FeCr₂O₄ produces:

(0.755 mol FeCr₂O₄) × (8 mol K₂CrO₄ / 4 mol FeCr₂O₄) = 1.51 mol K₂CrO₄

Calculate the theoretical mass of K₂CrO₄:

(1.51 mol K₂CrO₄) × (194.19 g K₂CrO₄ / 1 mol K₂CrO₄) = 293.227 g K₂CrO₄

%yield of K₂CrO₄ = (practical mass of K₂CrO₄ / theoretical mass of K₂CrO₄) × 100%

%yield of K₂CrO₄ = (194 g K₂CrO₄ / 293.227 g K₂CrO₄) × 100% = 66.16% = 66.2%

%yield of K₂CrO₄ = 66.2%

Answer: The percent yield of K₂CrO₄ is 66.2%.