Question

Ammonia gas combines with hydrogen chloride gas, forming solid ammonium chloride.

Answer 1

Answer:

a. The balanced chemical equation for the reaction

$NH_3(g)+HCl(g)\rightarrow NH_4Cl(g)$

b. HCl is the limiting reagent.

c. 7.3295 grams of ammonium chloride could form from the reaction mixture in part (b).

d. 0.6715 grams of ammonia is left over in the reaction mixture in part b.

Explanation:

a. The balanced chemical equation for the reaction

$NH_3(g)+HCl(g)\rightarrow NH_4Cl(g)$

b.

Moles of ammonia =$\frac{3.0 g}{17 g/mol}=0.1765 mol$

Moles of HCl = $\frac{5.0 g}{36.5 g/mol}=0.1370 mol$

According to reaction, 1 mole of HCl reacts with 1 mole of ammonia . Then 0.1370 moles of HCl will react with :

$\frac{1}{1}\times 0.1370 mol=0.1370 mol$ of ammonia

Hence, HCl is the limiting reagent.

c.

Since, HCl is a limiting reagent amount of ammonium chloride will depend upon moles of HCl.

According to reaction, 1 mole of HCl gives with 1 mole of ammonium chloride . Then 0.1370 moles of HCl will give :

$\frac{1}{1}\times 0.1370 mol=0.1370 mol$ of ammonium chloride

Mass of 0.1370 moles of ammonium chloride :

53.5 g/mol × 0.1370 mol =  7.3295 g

7.3295 grams of ammonium chloride could form from the reaction mixture in part (b).

d.

Moles of ammonia =$\frac{3.0 g}{17 g/mol}=0.1765 mol$

HCl is a limiting reagent and ammonia is an excessive reagent.

According to reaction, 1 mole of HCl reacts with 1 mole of ammonia . Then 0.1370 moles of HCl will react with :

$\frac{1}{1}\times 0.1370 mol=0.1370 mol$ of ammonia

Moles of Ammonia reacted = 0.1370 mol

Moles of ammonia left unreacted = 0.1765 mol - 0.1370 mol = 0.0395 mol

Mass of 0.0395 moles of ammonia :

0.0395 mol × 17 g/mol = 0.6715 g

0.6715 grams of ammonia is left over in the reaction mixture in part b.