Question

What mass of CaCO3 is required to remove 155 g of SO2 in the reaction 2 SO2(g) + 2 CaCO3(s) + O2(g) → 2 CaSO4(s) + 2 CO2(g)?

a) 122.5 g
b) 155 g
c) 310 g
d) 77.5 g

Answer 1

To remove 155 g of SO2, you would theoretically need 242.21 g of CaCO3, as determined by stoichiometric calculations. However, this result is not within the provided answer choices, suggesting a possible error in the question or given options.

Step-by-step explanation:

The calculation to determine the mass of CaCO3 needed to remove 155 g of SO2 involves stoichiometry. Firstly, we need to use the molar masses of SO2 and CaCO3 to convert the given mass of SO2 to moles:

Molar mass of SO2 = 64.066 g/mol

155 g SO2 x (1 mol SO2 / 64.066 g SO2) = 2.42 mol SO2

From the balanced equation, the mole ratio of SO2 to CaCO3 is 1:1, therefore 2.42 mol of SO2 will react with 2.42 mol of CaCO3. Next, we use the molar mass of CaCO3 to find the mass:

Molar mass of CaCO3 = 100.0869 g/mol

2.42 mol CaCO3 x (100.0869 g CaCO3 / 1 mol CaCO3) = 242.21 g CaCO3

Thus, the mass of CaCO3 required to remove 155 g of SO2 is found by rounding to the nearest half-value according to the choices given in the question, which is 242.5 g, but since this value is not an option provided, there could be a mistake in the question or the options and the correct answer from the ones provided cannot be conclusively determined.