Question

2. 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.

Answer 1

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* 2=44.8 L` and produce 2 moles of
`H_2SO_4`

mass of 2 moles of
`H_2SO_4=2* 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
`=(196)/(44.8)* {67.2}=294 g` of
`H_2SO_4`

mass of
`H_2SO_4`= 294 g

density of
`H_2SO_4=1.84gcm^(-3)`

`Volume=(mass)/(density)`

`Volume=(294 g)/(1.84gcm^(-3))`

Thus Volume of
`H_2SO_4=160ml=0.16L`