2 HNO3 + 1 Na2S -> H2S + 2 NaNO3
How many moles of H2S would we expect to be formed by reaction if 3.50 moles of HNO3 reacted completely?
3.50 mols HNO3 x (1 mol H2S/2 mols HNO3 = 3.50 x 1/2 = ?
To determine the number of moles of H2S formed in the reaction, we need to use the stoichiometry of the balanced chemical equation.
From the balanced equation: 2 HNO3 + 1 Na2S -> 1 H2S + 2 NaNO3
We can see that the molar ratio between HNO3 and H2S is 2:1. This means that for every 2 moles of HNO3, we expect to form 1 mole of H2S.
Given that we have 3.50 moles of HNO3, we can calculate the moles of H2S formed as follows:
Moles of H2S = (Moles of HNO3 / Ratio of HNO3 to H2S)
Moles of H2S = (3.50 moles / 2) = 1.75 moles
Therefore, we would expect 1.75 moles of H2S to be formed by reaction if 3.50 moles of HNO3 reacted completely.