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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.

Respuesta :

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

Part B: Volume of [tex]H_2SO_4[/tex]=0.16L

Explanation:

[tex]SO_2(g)+O_2(g)+H_2O(g)\rightarrow H_2SO_4(l)[/tex]

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: [tex]2SO_2(g)+O_2(g)+2H_2O(g)\rightarrow 2H_2SO_4(l)[/tex]

Part B: 1 mole of [tex]SO_2[/tex] occupies 22.4 L at STP

2 moles of [tex]SO_2[/tex] will occupy [tex]22.4\times 2=44.8 L[/tex] and produce 2 moles of [tex]H_2SO_4[/tex]

mass of 2 moles of [tex]H_2SO_4=2\times 98g=196g[/tex]

Thus 44.8 L of  [tex]SO_2[/tex] will produce 196 g of [tex]H_2SO_4[/tex]          

67.2 L of  [tex]SO_2[/tex] will produce [tex]=\frac{196}{44.8}\times {67.2}=294 g[/tex] of [tex]H_2SO_4[/tex]

mass of  [tex]H_2SO_4[/tex]= 294 g

density of [tex]H_2SO_4=1.84gcm^{-3}[/tex]

[tex]Volume=\frac{mass}{density}[/tex]

[tex]Volume=\frac{294 g}{1.84gcm^{-3}}[/tex]

Thus Volume of [tex]H_2SO_4=160ml=0.16L[/tex]





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