Question

use the following information to answer the next question
the unbalanced equation for the decomposition of ammonia gas is shown below.

1. at satp, the mass of ammonia required to produce 15l of hydrogen gas would be______g

(record your 3 - digit answer in the numeric response section of the answer sheet)
your answer

Explanation:

Step1: Balance the chemical equation

The balanced equation for the decomposition of ammonia is \(2NH_{3(g)}
ightarrow N_{2(g)}+3H_{2(g)}\)

Step2: Use the molar volume at STP

At STP (Standard Temperature and Pressure), the molar volume of a gas \(V_m = 22.4\ L/mol\).
First, find the number of moles of \(H_2\) gas. Given \(V = 15\ L\) of \(H_2\). Using the formula \(n=\frac{V}{V_m}\), we have \(n_{H_2}=\frac{15\ L}{22.4\ L/mol}\approx0.67\ mol\)

Step3: Use mole - ratio from the balanced equation

From the balanced equation \(2NH_{3(g)}
ightarrow N_{2(g)}+3H_{2(g)}\), the mole ratio of \(NH_3\) to \(H_2\) is \(\frac{n_{NH_3}}{n_{H_2}}=\frac{2}{3}\). So \(n_{NH_3}=\frac{2}{3}n_{H_2}\)
Substitute \(n_{H_2}\approx0.67\ mol\) into the equation: \(n_{NH_3}=\frac{2}{3}\times0.67\ mol\approx0.45\ mol\)

Step4: Calculate the mass of \(NH_3\)

The molar mass of \(NH_3\), \(M=(14 + 3\times1)\ g/mol=17\ g/mol\)
Using the formula \(m = n\times M\), where \(n = n_{NH_3}\approx0.45\ mol\) and \(M = 17\ g/mol\)
\(m=0.45\ mol\times17\ g/mol = 7.65\ g\)

Answer:

\(7.65\)