Question

При пропускании оксида серы(4) через раствор гидроксида кальция образуется гидросульфит кальция. Расчитайте объем оксида серы (4), взятого при нормальных условиях, если гидросульфита образовалось 101 кг.

Answer 1

Ca(OH)2+2SO2=Ca(HSO3)2

n(Ca(HSO3)2)=101000:202=500моль

n(SO2)=2*n(Ca(HSO3)2)=1000моль

V(SO2)=n*Vm=1000*22,4=22400дм3=22,4м3

 

Answer 2

Calculating the Volume of Sulfur Dioxide

When sulfur dioxide (SO2) is passed through a solution of calcium hydroxide (Ca(OH)2), calcium hydroxide is converted to calcium sulfite. To calculate the volume of sulfur dioxide taken under normal conditions, we can use the stoichiometry of the reaction and the given mass of calcium sulfite formed.

The balanced chemical equation for the reaction is: SO2 + Ca(OH)2 → CaSO3 + H2O

From the stoichiometry of the reaction, we can see that 1 mole of SO2 reacts with 1 mole of Ca(OH)2 to produce 1 mole of CaSO3. We can use the molar mass of CaSO3 to calculate the moles of CaSO3 formed, and then use the stoichiometry to find the moles of SO2 reacted.

Calculation Steps

1. Calculate the moles of CaSO3 formed from the given mass of 101 kg using its molar mass. 2. Use the stoichiometry of the reaction to find the moles of SO2 reacted. 3. Finally, calculate the volume of SO2 at normal conditions using the ideal gas law.

Calculation

1. Calculate moles of CaSO3 formed: - The molar mass of CaSO3 is 136.14 g/mol. - Moles of CaSO3 = Mass of CaSO3 / Molar mass of CaSO3 - Moles of CaSO3 = 101,000 g / 136.14 g/mol - Moles of CaSO3 = 741.97 mol

2. Use stoichiometry to find moles of SO2 reacted: - From the balanced chemical equation, 1 mole of SO2 produces 1 mole of CaSO3. - Moles of SO2 = Moles of CaSO3 - Moles of SO2 = 741.97 mol

3. Calculate the volume of SO2 at normal conditions: - Using the ideal gas law: PV = nRT - Assuming normal conditions are at 0°C and 1 atm pressure, R = 0.0821 L·atm/(K·mol) - V = (nRT) / P - V = (741.97 mol * 0.0821 L·atm/(K·mol) * 273 K) / 1 atm - V ≈ 16,961.6 L

The volume of sulfur dioxide taken at normal conditions is approximately 16,961.6 liters.