a reaction between ammonia gas and chlorine gas produces ammonium chloride and nitrogen gas as shown in the…

Explanation:

Step1: Calculate moles of reactants

The molar mass of $NH_3$ is $M_{NH_3}=14 + 3\times1= 17\ g/mol$, so the moles of $NH_3$, $n_{NH_3}=\frac{17.0\ g}{17\ g/mol}=1\ mol$. The molar mass of $Cl_2$ is $M_{Cl_2}=2\times35.5 = 71\ g/mol$, so the moles of $Cl_2$, $n_{Cl_2}=\frac{20.0\ g}{71\ g/mol}\approx0.282\ mol$.

Step2: Determine the limiting reactant

From the balanced - chemical equation $8NH_3(g)+3Cl_2(g)\rightarrow N_2(g)+6NH_4Cl(l)$, the mole - ratio of $NH_3$ to $Cl_2$ is $\frac{n_{NH_3}}{n_{Cl_2}}=\frac{8}{3}$. For the given amounts, if all $0.282\ mol$ of $Cl_2$ reacts, the moles of $NH_3$ required is $n_{NH_3\ required}=0.282\ mol\times\frac{8}{3}\approx0.752\ mol$. Since $1\ mol>0.752\ mol$, $Cl_2$ is the limiting reactant.

Step3: Calculate moles of $NH_4Cl$ produced

From the balanced equation, the mole - ratio of $Cl_2$ to $NH_4Cl$ is $\frac{n_{Cl_2}}{n_{NH_4Cl}}=\frac{3}{6}=\frac{1}{2}$. So the moles of $NH_4Cl$ produced, $n_{NH_4Cl}=2\times n_{Cl_2}=2\times0.282\ mol = 0.564\ mol$.

Step4: Calculate mass of $NH_4Cl$ produced

The molar mass of $NH_4Cl$ is $M_{NH_4Cl}=14 + 4\times1+35.5=53.5\ g/mol$. The mass of $NH_4Cl$ produced, $m_{NH_4Cl}=n_{NH_4Cl}\times M_{NH_4Cl}=0.564\ mol\times53.5\ g/mol\approx30.2\ g$.

Answer:

$30.2$