Question

1. how many grams of ammonium chloride are formed when 3.1 l of hydrogen chloride gas and 3.1 l of ammonia gas react? the density of the hydrogen chloride gas is 1.6 g/l and the density of ammonia is 0.77 g/l.

Explanation:

Step1: Calculate mass of ammonia

Use the formula $m =
ho V$, where $
ho$ is density and $V$ is volume. Given $
ho_{NH_3}=0.77\ g/L$ and $V = 3.1\ L$.
$m_{NH_3}=
ho_{NH_3}V=0.77\ g/L\times3.1\ L = 2.387\ g$

Step2: Write chemical - reaction equation

The reaction between ammonia ($NH_3$) and hydrogen chloride ($HCl$) is $NH_3(g)+HCl(g)
ightarrow NH_4Cl(s)$.

Step3: Calculate moles of ammonia

The molar mass of $NH_3$ is $M_{NH_3}=(14 + 3\times1)\ g/mol=17\ g/mol$. The number of moles of $NH_3$, $n_{NH_3}=\frac{m_{NH_3}}{M_{NH_3}}=\frac{2.387\ g}{17\ g/mol}\approx0.1404\ mol$.

Step4: Determine moles of ammonium chloride

From the balanced chemical equation, the mole - ratio of $NH_3$ to $NH_4Cl$ is 1:1. So, $n_{NH_4Cl}=n_{NH_3}\approx0.1404\ mol$.

Step5: Calculate mass of ammonium chloride

The molar mass of $NH_4Cl$ is $M_{NH_4Cl}=(14 + 4\times1+35.5)\ g/mol = 53.5\ g/mol$. Then $m_{NH_4Cl}=n_{NH_4Cl}M_{NH_4Cl}=0.1404\ mol\times53.5\ g/mol\approx7.51\ g$

Answer:

The mass of ammonia that reacts is $2.387\ g$ and the mass of ammonium chloride formed is approximately $7.51\ g$.