3NO₂(g) + H₂O(l) → 2HNO₃(l) + NO(g). If 3.9 mol NO₂ and 0.50 mol H₂O combine and react completely, how many moles of the excess reactant are present after the reaction is complete?

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3NO₂(g) + H₂O(l) → 2HNO₃(l) + NO(g). If 3.9 mol NO₂ and 0.50 mol H₂O combine and react completely, how many moles of the excess reactant are present after the reaction is complete?

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Answer 1 · 2015-04-10T16:24:21

2.4 mol of ${NO}_{2}$ $"NO"_2$ are present after the reaction is complete.

This is similar to an earlier question on Socratic, so I will simply go through the steps.

The balanced equation is

$3 {NO}_{2} + H_{2} O \to {2HNO}_{3} + NO$ $3"NO"_2+ "H"_2"O" → "2HNO"_3 + "NO"$

Step 1. Identify the limiting reactant.

${From NO}_{2} : 3.9 mol NO₂ \times \frac{1 mol NO}{3 mol NO₂} = 1.3 mol NO$ $"From NO"_2: 3.9 cancel("mol NO₂") × "1 mol NO"/(3 cancel("mol NO₂")) = "1.3 mol NO"$

${From H}_{2} O : 0.50 mol H₂O \times \frac{1 mol NO}{1 mol H₂O} = 0.50 mol NO$ $"From H"_2"O": 0.50 cancel("mol H₂O") × ("1 mol NO")/(1cancel("mol H₂O")) = "0.50 mol NO"$

$H_{2} O$ $"H"_2"O"$ is the limiting reactant, so ${NO}_{2}$ $"NO"_2$ is the excess reactant.

Step 2. Calculate the moles of excess reactant used up.

${Moles of NO}_{2} used = 0.50 mol H₂O \times \frac{{3 mol NO}_{2}}{1 mol H₂O} = {1.50 mol NO}_{2}$ $"Moles of NO"_2" used" = 0.50 cancel("mol H₂O") × ("3 mol NO"_2)/(1 cancel("mol H₂O")) = "1.50 mol NO"_2$

Step 3. Calculate the moles of excess reactant left over.

${Moles of excess NO}_{2} = 3.9 mol - 1.50 mol = 2.4 mol$ $"Moles of excess NO"_2 = "3.9 mol – 1.50 mol" = "2.4 mol"$

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3NO₂(g) + H₂O(l) → 2HNO₃(l) + NO(g). If 3.9 mol NO₂ and 0.50 mol H₂O combine and react completely, how many moles of the excess reactant are present after the reaction is complete?

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