Question

multiple choice 1 point
for the reaction represented by the equation so₃ + h₂o → h₂so₄, calculate the percentage yield if 500. g of sulfur trioxide react with excess water to produce 575 g of sulfuric acid in lab.
93.9%
88.3%
91.2%

Explanation:

Step1: Calculate the molar mass of \(SO_3\) and \(H_2SO_4\)

The molar mass of \(SO_3\): \(M_{SO_3}=32+(16\times3)=80\space g/mol\)
The molar mass of \(H_2SO_4\): \(M_{H_2SO_4}=(1\times2)+32+(16\times4)=98\space g/mol\)

Step2: Calculate the moles of \(SO_3\)

Moles of \(SO_3\), \(n_{SO_3}=\frac{m_{SO_3}}{M_{SO_3}}=\frac{500\space g}{80\space g/mol}=6.25\space mol\)

Step3: Calculate the theoretical yield of \(H_2SO_4\)

From the balanced equation \(SO_3 + H_2O
ightarrow H_2SO_4\), the mole ratio of \(SO_3\) to \(H_2SO_4\) is \(1:1\). So moles of \(H_2SO_4\) (theoretical) \(n_{H_2SO_4}^{theo}=6.25\space mol\)
Theoretical mass of \(H_2SO_4\), \(m_{H_2SO_4}^{theo}=n_{H_2SO_4}^{theo}\times M_{H_2SO_4}=6.25\space mol\times98\space g/mol = 612.5\space g\)

Step4: Calculate the percentage yield

Percentage yield \(=\frac{m_{H_2SO_4}^{actual}}{m_{H_2SO_4}^{theo}}\times100\%=\frac{575\space g}{612.5\space g}\times100\% = 93.9\%\)

Answer:

93.9%