3NO2(g) + H2O(l) → 2HNO3(l) + NO(g)
How many moles of nitric acid are produced starting from 5.00 moles of NO2(g) and excess water?
mols HNO3/ mols NO2 = 2/3
so (2/3)5 = 10/3
To determine the number of moles of nitric acid produced, you need to use the balanced chemical equation provided and the given number of moles of NO2.
The balanced chemical equation is:
3NO2(g) + H2O(l) → 2HNO3(l) + NO(g)
From the equation, you can see that 3 moles of NO2 react with 1 mole of H2O to produce 2 moles of HNO3 and 1 mole of NO.
Given that you have 5.00 moles of NO2, you can set up a proportion using the coefficients from the balanced equation:
(3 moles NO2 / 3 moles NO2) = (2 moles HNO3 / X moles HNO3)
Cross-multiplying and solving for X (moles of HNO3), you get:
X = (2 moles HNO3 * 5.00 moles NO2) / 3 moles NO2
X = 10.00 moles HNO3 / 3 moles NO2
Calculating this expression, you find:
X ≈ 3.33 moles HNO3
Therefore, approximately 3.33 moles of nitric acid are produced starting from 5.00 moles of NO2 and excess water.