Sulfur dioxide obtained during pyrite roasting was passed through a sodium hydroxide

Sulfur dioxide obtained during pyrite roasting was passed through a sodium hydroxide solution weighing 170 g to determine the volume of gas

To solve, we write down the equations:
4FeS2 + 11O2 = 8SO2 + 2Fe2O3 – pyrite roasting, sulfur oxide (sulfur dioxide) is released;
SO2 + 2NaOH = Na2SO3 + H2O – ion exchange.
Let’s carry out the calculations:
M (SO2) = 64 g / mol;
M (NaOH) = 39.9 g / mol.
Determine the number of moles of sodium hydroxide, if the mass is known:
Y (NaOH) = m / M = 170 / 39.9 = 4.26 mol.
Let’s make a proportion according to the equation:
X mol (SO2) – 4.26 mol (NaOH);
-1 mol – 2 mol hence, X mol (SO2) = 1 * 4.26 / 2 = 2.13 mol.
Let’s calculate the volume of sulfur dioxide:
V (SO2) = 2.13 * 22.4 = 47.71 liters.
Answer: The volume of sulfur dioxide is 47.71 liters.