Question

supposing there are 1.50 x 10^6 g of sulfur dioxide emitted by an industrial plant , how many moles of calcium carbonate and oxygen that are needed to remove the given amount of pollutant ? spacecrafts and submarines use LiOH to capture exhaled CO2 in the reaction producing lithium carbonate and water vapor .​

Answer 1

2SO₂ + 2CaCO₃ + O₂ → 2CaSO₄ + 2CO₂

​M(SO₂) = 64.06 g/mol

n(SO₂) $= \frac{m}{M} = \frac{1.50 \times 10^6}{64.06} = 23.41 \times 10^3 \; mol$

n(CaCO₃) $= n(SO_2) = 23.41 \times 10^3 \; mol$

n(O₂) $= \frac{1}{2}n(SO_2) = \frac{23.41 \times 10^3}{2} = 11.70 \times 10^3 \; mol$

Answer: $23.41 \times 10^3 \; mol \;and \;11.70 \times 10^3 \; mol$