Nitrogen dioxide reacts with water to form nitric acid and nitrogen monoxide according to the equation: 3NO2 (g) + H2O (l) →2HNO3 (l) + NO (g) Suppose that 26 mol NO2 and 7 mol H2O combine and react completely. How many moles of the reactant in excess are present after the reaction has completed?

Question

Jaylan Mccarthy

Chemistry

9 November, 04:34

Nitrogen dioxide reacts with water to form nitric acid and nitrogen monoxide according to the equation: 3NO2 (g) + H2O (l) →2HNO3 (l) + NO (g) Suppose that 26 mol NO2 and 7 mol H2O combine and react completely. How many moles of the reactant in excess are present after the reaction has completed?

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Kathleen · Answer 1

5 moles of NO₂ will remain after the reaction is complete

Explanation:

We state the reaction:

3NO₂ (g) + H₂O (l) → 2HNO₃ (l) + NO (g)

3 moles of nitric oxide can react with 1 mol of water. Ratio is 3:1, so we make this rule of three:

If 3 moles of nitric oxide need 1 mol of water to react

Then, 26 moles of NO₂ may need (26.1) / 3 = 8.67 moles of H₂O

We have 7 moles of water but we need 8.67 moles, so water is the limiting reactant because we do not have enough. In conclusion, the oxide is the reagent in excess. We can verify:

1 mol of water needs 3 moles of oxide to react

Therefore, 7 moles of water will need (7.3) / 1 = 21 moles of oxide

We have 26 moles of NO₂ and we need 21, so we still have oxide after the reaction is complete. We will have (26-21) = 5 moles of oxide that remains