Question

A) What is the limiting reactant when 200.0 grams of Fe2O3 reacts with 300.0 grams of C?

A) Fe2O3
B) C
C) CO2
D) Fe
b) How many moles of Fe are produced?

A) 200.0 moles
B) 400.0 moles
C) 600.0 moles
D) 800.0 moles
c) How many liters of CO2 gas at STP are produced?

A) 22.4 liters
B) 33.6 liters
C) 44.8 liters
D) 56.0 liters
d) Determine the mass of the excess reactant that would be left over after the reaction is completed.

A) 0 grams
B) 100.0 grams
C) 150.0 grams
D) 200.0 grams

Answer 1

The limiting reactant when 200.0 grams of Fe2O3 reacts with 300.0 grams of C is iron(III) oxide (Fe2O3), determined by the stoichiometric calculations based on the molar masses and the balanced chemical equation.

Step-by-step explanation:

To determine the limiting reactant when 200.0 grams of iron(III) oxide (Fe2O3) reacts with 300.0 grams of carbon (C), we must first convert these masses to moles. The molar mass of Fe2O3 is 159.7 grams per mole and the molar mass of C is 12.01 grams per mole. Using these molar masses:

• 200.0 g Fe2O3 × (1 mol / 159.7 g) = 1.253 moles of Fe2O3
• 300.0 g C × (1 mol / 12.01 g) = 24.98 moles of C

The balanced chemical equation for the reaction is:

Fe2O3(s) + 3C(s) → 2Fe(l) + 3CO(g)

Now, divide the moles of each reactant by its coefficient in the balanced equation:

• 1.253 moles Fe2O3 / 1 = 1.253
• 24.98 moles C / 3 = 8.327

Since the ratio of Fe2O3 is less, Fe2O3 is the limiting reactant. This means that all the Fe2O3 will be consumed in the reaction while some amount of C will be left unreacted.