Question

Nitrogen and hydrogen combine at a high temperature, in the presence of a catalyst, to produce ammonia.

N2(g)+3H2(g)⟶2NH3(g)

Assume 0.290 mol N2 and 0.954 mol H2 are present initially.

After complete reaction, how many moles of ammonia are produced?

Answer 1

17 reactions

Step-by-step explanation:

Answer 2

After performing a stoichiometry calculation using the provided moles of nitrogen and hydrogen, 0.580 moles of ammonia are produced after a complete reaction.

When nitrogen and hydrogen react to produce ammonia, the reaction follows a stoichiometric ratio according to the balanced chemical equation:

`N_(2) (g) + 3 H_(2)(g)` →
`2NH_(3) (g)`

To solve for the moles of ammonia produced, we must first determine the limiting reactant. We are given 0.290 mol
`N_(2)` and 0.954 mol
`H_(2)`. The stoichiometric ratio of hydrogen to nitrogen is 3:1 from the balanced equation; thus, to fully react with 0.290 mol of nitrogen, we would need 0.290 mol x 3 = 0.870 mol of hydrogen.

Since 0.954 mol of hydrogen is provided, which is more than needed, nitrogen is the limiting reactant.

Every 1 mol of
`N_(2)` produces 2 mol of
`NH_(3)`, so 0.290 mol of
`N_(2)` will produce 0.290 mol x 2 = 0.580 mol of
`NH_(3)`. Therefore, after the complete reaction, 0.580 moles of ammonia are produced.