Final answer:
To determine the mass of CaCO₃ required to react with 155 g of O₂, you must convert the mass of O₂ to moles, use the mole ratio from the balanced equation to find the moles of CaCO₃ needed, and then convert this to grams. The result is 969.84 g of CaCO₃.
Step-by-step explanation:
Stoichiometry Calculation to Determine Mass of CaCO₃
To determine the mass of calcium carbonate (CaCO₃) required to react with 155 g of O₂, we use stoichiometry based on the balanced chemical equation:
2SO₂(g) + 2CaCO₃(s) + O₂(g) ⇒ 2CaO₄(s) + 2CO₂(g)
Firstly, convert the mass of O₂ to moles using its molar mass (32.00 g/mol). Then, we apply the mole ratio between O₂ and CaCO₃ from the balanced equation.
Finally, convert the moles of CaCO₃ to mass using its molar mass. Conservation of mass also dictates that the mass of reactants equals the mass of products, thus reinforcing the stoichiometric calculations.
Step 1:
Calculate moles of O₂:
155 g O₂ × (1 mol O₂ / 32.00 g O₂) = 4.84375 mol O₂
Step 2:
Use the mole ratio 1 mol O₂ : 2 mol CaCO₃ to find moles of CaCO₃.
4.84375 mol O₂ × (2 mol CaCO₃ / 1 mol O₂) = 9.6875 mol CaCO₃
Step 3:
Convert moles of CaCO₃ to mass:
9.6875 mol CaCO₃ × (100.09 g CaCO₃ / 1 mol CaCO₃) = 969.84 g CaCO₃
Therefore, 969.84 g of CaCO₃ is required to react with 155 g of O₂.