Answer:
91.59%
Explanations
Given the reaction between sulfur trioxide and water expressed as:
[tex]SO_3(g)+H_2O(l)\rightarrow H_2SO_4(aq)_[/tex]
Determine the moles of SO3 and H2O
[tex]\begin{gathered} moles\text{ of SO}_3=\frac{mass}{molar\text{ mass}} \\ moles\text{ of SO}_3=\frac{61.5g}{80.06g\text{/mol}} \\ moles\text{ of SO}_3=0.768moles \end{gathered}[/tex]
For H2O
[tex]\begin{gathered} moles\text{ of H}_2O=\frac{11.2}{18} \\ moles\text{ of H}_2O=0.6222moles \end{gathered}[/tex]
Since the mole of water is lower than that of SO3, water will be the limiting reactant
According to stoichiometry, 1 mole of water produce 1 mole of H2SO4, hence the moles of H2SO4 required will be 0.6222moles
Determine the mass of H2SO4 (theoretical yield)
[tex]\begin{gathered} Mass\text{ of H}_2SO_4=mole\times molar\text{ mass} \\ Mass\text{ of H}_2SO_4=0.6222\times98.079 \\ Mass\text{ of }H_2SO_4=61.03grams \end{gathered}[/tex]
Determine the percentage yield of the reaction
[tex]\begin{gathered} \%yield=\frac{actual}{theoretical}\times100\% \\ \%yield=\frac{55.9}{61.03}\times100 \\ \%yield=\frac{5590}{61.03} \\ \%yield=91.59\% \end{gathered}[/tex]
Hence the percentage yield of the reaction is 91.59%
