Question

specific mass conversions quick check
nitrogen gas and hydrogen gas react to produce ammonia according to the following equation.
$\ce{n_{2} + 3h_{2} \to 2nh_{3}}$
the atomic mass of nitrogen is $14\\ g/mol$. the atomic mass of hydrogen is $1\\ g/mol$. when the reaction produces 68 grams of ammonia, how many grams of nitrogen were present initially?
(1 point)
○ 112
○ 56
○ 4
○ 12

Explanation:

Step1: Calculate the molar mass of ammonia ($\ce{NH3}$)

The molar mass of $\ce{NH3}$ is $14+(1\times3)=17\ \text{g/mol}$.

Step2: Find the number of moles of ammonia produced

Number of moles of $\ce{NH3}$, $n=\frac{m}{M}=\frac{68\ \text{g}}{17\ \text{g/mol}} = 4\ \text{mol}$.

Step3: Use the stoichiometry of the reaction

From the balanced equation $\ce{N2 + 3H2
ightarrow2NH3}$, the mole ratio of $\ce{N2}$ to $\ce{NH3}$ is $1:2$. So, if $n(\ce{NH3}) = 4\ \text{mol}$, then $n(\ce{N2})=\frac{4\ \text{mol}}{2}=2\ \text{mol}$.

Step4: Calculate the mass of nitrogen

The molar mass of $\ce{N2}$ is $14\times2 = 28\ \text{g/mol}$. Mass of $\ce{N2}$, $m=n\times M=2\ \text{mol}\times28\ \text{g/mol}=56\ \text{g}$.

Answer:

56