Combustion of coal releases sulfur dioxide into the atmosphere. The following pr

Question

Combustion of coal releases sulfur dioxide into the atmosphere. The following process converts this gas into sulfuric acid, a component of acid rain.
2SO2(g) + O2(g) 2SO3(g)
SO3(g) + H2O(l) H2SO4(aq)
If each tonne of coal produces 8.00 × 103 L of sulfur dioxide (measured at STP), what mass of sulfuric acid can result from combustion of each tonne of coal? (1 tonne = 1000 kg)

1.76 × 104 kg H2SO4

7.01 × 101 kg H2SO4

3.50 × 107 kg H2SO4

1.75 × 101 kg H2SO4

3.50 × 101 kg H2SO4

1.76 × 104 kg H2SO4

7.01 × 101 kg H2SO4

3.50 × 107 kg H2SO4

1.75 × 101 kg H2SO4

3.50 × 101 kg H2SO4

Explanation / Answer

First calculate the number of moles of SO2 in 8.00 x10^3 L at STP as follows:

1 mole of any gas at STP occupies 22.4 liters of volume.

By the use of this information, calculate the number of mole of SO2 as follows:

the number of mole of SO2 = total volume / molar volume

the number of mole of SO2 = 8.00 × 103 L / 22.4 liters / mol

the number of mole of SO2 =357.14 mol

Now use stoichiometry to solve for moles H2SO4:
moles H2SO4/1 tonne coal

=(357.14 mol moles SO2)*(2mol SO3/2 mol SO2)*(1 mol H2SO4/1 mol SO3)

357.14 mol of H2SO4

Now calculate the for mass of 357.14 mol H2SO4 by using its molar mass of H2SO4:

molar mass H2SO4 = 98.079 g/mol

mass of H2SO4/1 tonne coal

=357.14 mol of H2SO4* molar mass H2SO4

mass of H2SO4/1 tonne coal

=357.14 mol of H2SO4* 98.079 g/mol H2SO4

= 3.50x10^4 g or 3.50x10^1 k g

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