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  1. Sulfur dioxide gas (SO2) reacts with excess oxygen gas (O2) and excess liquid water (H2O) to form liquid sulfuric acid (H2SO4) in the UNBALANCED equation below: SO2 + O2 + H2O  H2SO4 In the laboratory, a chemist carries out this reaction at STP with 67.2 L of sulfur dioxide (SO2). How many liters of H2SO4 did the chemist produce? 1 mole of ANY gas = 22.4 L of that same gas at STP • Part A: Write a balanced equation for the reaction. • Part B: Calculate the number of liters of H2SO4 produced.

Verified Answer :

Answer: Part A: \(2SO_2(g)+O_2(g)+2H_2O(g)\rightarrow 2H_2SO_4(l)\)

Part B: Volume of \(H_2SO_4\)=0.16L

Explanation:

\(SO_2(g)+O_2(g)+H_2O(g)\rightarrow H_2SO_4(l)\)

According to law of conservation of mass, the atoms on product side must be equal to the atoms on reactant side so that the mass remains conserved.

Part A: The balanced chemical reaction is: \(2SO_2(g)+O_2(g)+2H_2O(g)\rightarrow 2H_2SO_4(l)\)

Part B: 1 mole of \(SO_2\) occupies 22.4 L at STP

2 moles of \(SO_2\) will occupy \(22.4\times 2=44.8 L\) and produce 2 moles of \(H_2SO_4\)

mass of 2 moles of \(H_2SO_4=2\times 98g=196g\)

Thus 44.8 L of  \(SO_2\) will produce 196 g of \(H_2SO_4\)          

67.2 L of  \(SO_2\) will produce \(=\frac{196}{44.8}\times {67.2}=294 g\) of \(H_2SO_4\)

mass of  \(H_2SO_4\)= 294 g

density of \(H_2SO_4=1.84gcm^{-3}\)

\(Volume=\frac{mass}{density}\)

\(Volume=\frac{294 g}{1.84gcm^{-3}}\)

Thus Volume of \(H_2SO_4=160ml=0.16L\)





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