3NO2(g) + H2O(l) → 2HNO3(l) + NO(g)



How many moles of NO2(g) are required to react with 1.00 mole of water to produce 3.00 moles of nitric acid?

Question 24 options:

6.00 mol


1.50 mol


2.00 mol


Not enough water is available to produce 3.00 moles of nitric acid.


4.50 mo

2.00 mol

To find the number of moles of NO2(g) required to react with 1.00 mole of water to produce 3.00 moles of nitric acid, we first need to determine the mole ratio between NO2(g) and H2O(l) in the balanced chemical equation.

The balanced equation is:
3NO2(g) + H2O(l) → 2HNO3(l) + NO(g)

From the balanced equation, we can see that the mole ratio between NO2(g) and H2O(l) is 3:1. This means that for every 3 moles of NO2(g), 1 mole of H2O(l) is required.

Given that we have 1.00 mole of water, we can set up a proportion to find the number of moles of NO2(g) required:
3 mol NO2(g) / 1 mol H2O(l) = x mol NO2(g) / 1.00 mol H2O(l)

Simplifying the proportion gives us:
3 mol NO2(g) = x mol NO2(g)

So, the number of moles of NO2(g) required is 3.00 moles. Therefore, the correct answer is 3.00 mol.