Question

If 22.27 grams of Fe2O3 is mixed with 8.080 grams of CO and allowed to react according to the balanced equation: Fe2O3(s) + 3 CO(g) → 2 Fe(s) + 3 CO2(g) What is the limiting reagent? COFe2O3 How many grams of CO2 could be produced? grams CO2 What mass, in grams, of the excess reagent will remain?

Answer 1

The answer to your question is

a) Limiting reactant = CO

b) 12,7 g of CO₂

c) 6.9 g of Fe₂O₃

Step-by-step explanation:

Data

Fe₂O₃ = 22.27 g

CO = 8.08 g

Molecular weight Fe₂O₃ = 112 + 48 = 160 g

CO = 12 + 16 = 28 g

CO₂ = 12 + 32 = 44 g

Reaction

Fe₂O₃ + 3 CO ⇒ 2 Fe + 3 CO₂

Limiting reactant

Calculate the theoretical and experimental proportion

Theoretical Fe2O3 / 3CO = 160 / (3x 28) = 160 / 84 = 1.9

Experimental Fe2O3 / CO = 22.27 / 8.08 = 2.7

The limiting reactant is CO because the experimental proportion was higher than the theoretical proportion.

Amount of CO2

3(28) g of CO -------------- 3 (44) g of CO₂

8.08 g of CO -------------- x

x = (8.08 x 3 x 44) / 3(28)

x = 1066.6 / 84

x = 12.69 g of CO₂

Excess reactant

3(28) g of CO --------------- 160 g of Fe₂O₃

8.08 g of CO ----------------- x

x = (8.08 x 160) / 3(28)

x = 15.4 g of Fe₂O₃

Excess reactant = 22.27 - 15.4

= 6.87 g of Fe₂O₃