Question

Which reactant is the limiting reactant and which is the excess reactant when there are 15.0 g of hydrogen gas and 10.0 g of nitrogen gas?

3H2(g) + N2(g) → 2NH3(g)

Answer 1

To determine the limiting reactant and excess reactant, compare the moles of each reactant to the stoichiometric ratio in the balanced equation. In this case, hydrogen gas is the limiting reactant and nitrogen gas is the excess reactant.

Step-by-step explanation:

In order to determine the limiting reactant and excess reactant, we need to compare the molar amounts of each reactant provided with their stoichiometric ratios in the balanced equation. For example, if there are 15.0 g of hydrogen gas (H2) and 10.0 g of nitrogen gas (N2) given in the question, we can convert these masses to moles using their molar masses.

Afterwards, we compare the mole ratios of H2 and N2 with the stoichiometric ratio in the balanced equation (3H2 + N2 → 2NH3). The reactant that yields the smaller amount of product is the limiting reactant, while the other reactant is the excess reactant. In this case, the limiting reactant is hydrogen gas (H2) and nitrogen gas (N2) is the excess reactant.