5.0mol of Fe2O3(s) and 6.0mol of CO(g) react according to the equation below. What is the limiting reactant and how many moles of the excess reactant remain unreacted?Fe2O3(s) + 3CO(g) → 2Fe(s) + 3CO2(g)
Question
5.0mol of Fe2O3(s) and 6.0mol of CO(g) react according to the equation below. What is the limiting reactant and how many moles of the excess reactant remain unreacted?Fe2O3(s) + 3CO(g) → 2Fe(s) + 3CO2(g)
Solution
First, let's identify the stoichiometry of the reaction. According to the balanced chemical equation, 1 mole of Fe2O3 reacts with 3 moles of CO to produce 2 moles of Fe and 3 moles of CO2.
Next, let's determine which reactant is the limiting reactant. This is the reactant that will be completely consumed first and will determine the amount of product formed.
We have 5.0 moles of Fe2O3 and 6.0 moles of CO. According to the stoichiometry of the reaction, 1 mole of Fe2O3 requires 3 moles of CO. Therefore, 5.0 moles of Fe2O3 would require 5.0 * 3 = 15.0 moles of CO.
However, we only have 6.0 moles of CO. This means that CO is the limiting reactant because we don't have enough of it to react with all the Fe2O3.
Now, let's calculate how many moles of the excess reactant (Fe2O3) remain unreacted.
We know that 1 mole of Fe2O3 reacts with 3 moles of CO. So, 6.0 moles of CO would react with 6.0 / 3 = 2.0 moles of Fe2O3.
We initially had 5.0 moles of Fe2O3. So, the amount of Fe2O3 that remains unreacted is 5.0 - 2.0 = 3.0 moles.
So, the limiting reactant is CO and 3.0 moles of Fe2O3 remain unreacted.
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