Question

question 11
the fertilizer ammonium sulfate, (nh4)2so4, is prepared by reacting ammonia and sulfuric acid together according to the following reaction:
2nh3(g)+h2so4(aq)→(nh4)2so4(aq)
how many kilograms of nh3 are needed to produce 2.60×105 kg of (nh4)2so4?
2.60×105 kg nh3 needed
5.12×105 kg nh3 needed
6.70×104 kg nh3 needed
1.97×106 kg nh3 needed

Explanation:

Step1: Calculate the molar mass of \((NH_4)_2SO_4\) and \(NH_3\)

The molar mass of \((NH_4)_2SO_4\):

$$
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The molar mass of \(NH_3\):

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Step2: Convert the mass of \((NH_4)_2SO_4\) to moles

Given \(m_{(NH_4)_2SO_4}=2.60\times 10^{5}\space kg = 2.60\times 10^{8}\space g\)
The number of moles of \((NH_4)_2SO_4\), \(n_{(NH_4)_2SO_4}=\frac{m_{(NH_4)_2SO_4}}{M_{(NH_4)_2SO_4}}=\frac{2.60\times 10^{8}\space g}{132\space g/mol}\approx 1.97\times 10^{6}\space mol\)

Step3: Use the stoichiometry of the reaction

From the balanced equation \(2NH_3(g)+H_2SO_4(aq)\to(NH_4)_2SO_4(aq)\), the mole ratio of \(NH_3\) to \((NH_4)_2SO_4\) is \(2:1\)
So the number of moles of \(NH_3\), \(n_{NH_3} = 2\times n_{(NH_4)_2SO_4}=2\times1.97\times 10^{6}\space mol = 3.94\times 10^{6}\space mol\)

Step4: Convert moles of \(NH_3\) to mass

The mass of \(NH_3\), \(m_{NH_3}=n_{NH_3}\times M_{NH_3}=3.94\times 10^{6}\space mol\times17\space g/mol = 6.70\times 10^{7}\space g=6.70\times 10^{4}\space kg\)

Answer:

\(6.70\times 10^{4}\space kg\space NH_3\) needed (the third option)