Question

a chemist measures the energy change \\( \delta h \\) during the following reaction:

\\( 2 \mathrm { nh } _ { 3 } ( g ) \
ightarrow \mathrm { n } _ { 2 } ( g ) + 3 \mathrm { h } _ { 2 } ( g ) \\) \\( \delta h = 160. \mathrm { kj } \\)

use the information to answer the following question.

suppose \\( 12.8 \mathrm { g } \\) of \\( \mathrm { nh } _ { 3 } \\) react.

calculate how much heat will be released or absorbed.

be sure your answer has the correct number of significant digits.

Explanation:

Step1: Calculate the molar mass of \(NH_3\)

The molar mass of \(N = 14.0\space g/mol\) and \(H=1.0\space g/mol\). For \(NH_3\), \(M=(14.0 + 3\times1.0)\space g/mol=17.0\space g/mol\)

Step2: Calculate the number of moles of \(NH_3\)

Using the formula \(n=\frac{m}{M}\), where \(m = 12.8\space g\) and \(M = 17.0\space g/mol\). So \(n=\frac{12.8\space g}{17.0\space g/mol}\approx0.753\space mol\)

Step3: Relate moles of \(NH_3\) to \(\Delta H\)

From the reaction \(2NH_3(g)\to N_2(g)+3H_2(g)\), \(\Delta H = 160\space kJ\) for \(2\space mol\) of \(NH_3\). Let \(x\) be the heat change for \(0.753\space mol\) of \(NH_3\). Using proportion \(\frac{x}{0.753\space mol}=\frac{160\space kJ}{2\space mol}\)

$$x=\frac{160\space kJ\times0.753\space mol}{2\space mol}$$

$$x = 60.24\space kJ$$

Answer:

\(60.2\space kJ\) (rounded to three significant digits)