Question

1. determine the amount of heat, in kj, given off when 1.26 × 10⁴ g of ammonia are produced according to the equation below. (molar mass nh₃ = 17.0 g/mol) n₂(g) + 3 h₂(g) → 2 nh₃(g) δh° = -92.6 kj/mol a. 3.43 × 10⁴ kj b. 6.86 × 10⁴ kj c. 7.93 × 10⁵ kj d. 1.32 × 10⁴ kj e. 2.55 × 10⁵ kj

Explanation:

Step1: Calculate the number of moles of \(NH_3\)

Use the formula \(n=\frac{m}{M}\), where \(m = 1.26\times10^{4}\space g\) and \(M = 17.0\space g/mol\).
\(n=\frac{1.26\times 10^{4}\space g}{17.0\space g/mol}\)

Step2: Relate moles of \(NH_3\) to heat change

From the equation \(N_{2}(g)+3H_{2}(g)\to2NH_{3}(g)\space\Delta H^{\circ}=- 92.6\space kJ/mol\), for \(2\space mol\) of \(NH_3\) produced, the heat change is \(-92.6\space kJ\).
Let \(x\) be the heat change for \(n\) moles of \(NH_3\). Then \(x=\frac{-92.6\space kJ}{2\space mol}\times n\)

Substitute \(n=\frac{1.26\times 10^{4}}{17.0}\) into the heat - change formula:

$$
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$$

The negative sign indicates heat is given off.

Answer:

A. \(3.43\times 10^{4}\space kJ\)