Question

question 11
the fertilizer ammonium sulfate, (nh₄)₂so₄, is prepared by reacting ammonia and sulfuric acid together according to the following reaction:
2 nh₃(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 \(NH_3\) and \((NH_4)_2SO_4\)

The molar mass of \(NH_3\) (\(M_{NH_3}\)): \(M_{NH_3}=14 + 3\times1=17\space g/mol\)
The molar mass of \((NH_4)_2SO_4\) (\(M_{(NH_4)_2SO_4}\)): \(M_{(NH_4)_2SO_4}=2\times(14 + 4\times1)+32+4\times16=132\space g/mol\)

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 \((NH_4)_2SO_4\), \(n_{(NH_4)_2SO_4}=\frac{m_{(NH_4)_2SO_4}}{M_{(NH_4)_2SO_4}}=\frac{2.60\times 10^{8}\space g}{132\space g/mol}\)

Step3: Use the stoichiometry of the reaction

From the reaction \(2NH_3(g)+H_2SO_4(aq)\to(NH_4)_2SO_4(aq)\), the mole ratio of \(NH_3\) to \((NH_4)_2SO_4\) is \(n_{NH_3}:n_{(NH_4)_2SO_4}=2:1\)
So \(n_{NH_3} = 2\times n_{(NH_4)_2SO_4}=2\times\frac{2.60\times 10^{8}\space g}{132\space g/mol}\)

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

\(m_{NH_3}=n_{NH_3}\times M_{NH_3}=2\times\frac{2.60\times 10^{8}\space g}{132\space g/mol}\times17\space g/mol\)
\(m_{NH_3}=\frac{2\times2.60\times 10^{8}\times17}{132}\space g\approx 6.70\times 10^{7}\space g = 6.70\times 10^{4}\space kg\)

Answer:

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