Question

Sodium hydroxide reacts with carbondioxide as follows: 2 naoh(s) + co2 (g) → na2co3 (s) + h2o(l) which reagent is the limiting reactant when 1.85 mol of sodium hydroxide and 1.00 mol carbondioxide are allowed to react? how many moles of sodium carbonate can be produced? how many moles of the excess reactant remain after the completion of the reaction?

Answer 1

1) Balanced chemical equation:

2 NaOH(s) + CO₂ (g) → Na₂CO₃ (s) + H₂O (l)

2) Mole ratio

2 mole NaOH : 1 mol CO₂ : 1 mol Na₂CO₃ : 1 mol H₂O

3) Limiting reactant

1.85 mol NaOH / 1.00 mol CO₂ < 2 mole NaOH / 1 mol CO₂ ⇒ NaOH is not enough to react with 1.00 mole of CO₂ (more NaOH is needed), so NaOH is the limiting reactant.

Answer: the limiting reactant is NaOH

4) How many moles of sodium carbonate can be produced?

Set the proportion relation using the limiting reactant (the one that reacts completely) and the theoretical mole ratio.

2 mole NaOH / 1mol Na₂CO₃ = 1.85 mol NaOH / x

x = 0.925 mol Na₂CO₃

Answer: 0.925 moles Na₂CO₃

5) How many moles of the excess reactant remain after the completion of the reaction?

The excess reactant is CO₂.

The amount of CO₂ consumed is calculated with a proportion:

2 mol NaOH / 1 mol CO₂ = 1.85 mol NaOH / x

⇒ x = 1.85 / 2 = 0.925mol CO₂

The amount remaining is the original amount less the amount that reacted = 1 mol - 0.925 mol = 0.075 mol.

Answer: 0.075 mol CO₂