Question

How many moles of which reactant will remain if 1.39 moles of N2 and 3.44 moles of
H2 will react to form ammonia? find out how many grams of ammonia can be formed
and how many moles of limiting reactant is required to completely exhaust the other
reactant that is in excess?

Answer 1

#74962

$$N_2 + 3H_2 \longrightarrow 2NH_3$$ $$1.39 \quad 3.44 \quad \text{①} \quad \text{③-④}$$

Limiting reagent

ACC. to stoichiometric Eqn

$$\begin{array}{l} 1N_2 \rightarrow 3H_2 \\ 1.39 \rightarrow 3 \times 1.39 = 4.17 \text{ moles} \end{array}$$

So, limiting reagent is $(N_2)$.

$$3 \text{ molecules } \gamma N_2 \rightarrow 2 \text{ molecules } \gamma NH_3$$ $$1 \quad \dots \quad \dots \quad \rightarrow \quad \frac{2}{3}$$ $$3.44 \quad \dots \quad \dots \quad \rightarrow \quad \frac{2}{3} \times 3.44$$ $$= 2.293 \text{ moles}$$ $$\begin{array}{l} \text{Now, moles in grams } \gamma NH_3 \text{ is:} \\ = 2.293 \times 17 \quad \text{[: Molecular Mass } \gamma NH_3 = 17\text{]} \\ = 38.981 \text{ g} \end{array}$$

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