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

Answer:

The answer to your question is

a) Limiting reactant = CO

b) 12,7 g of CO₂

c) 6.9 g of Fe₂O₃

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