5.00 g of copper(II) carbonate decomposes to form copper(II) oxide and carbon dioxide: CuCO3(g) → CuO(s) + CO2(g). Calculate the maximum mass of carbon dioxide that can be produced. (Mr of CuCO3 = 123.5, Mr of CO2 = 44.0)

Moles of CuCO3 (using triangle) = 5.00/123.5 = 0.0450 mol.1 mol of CuCO3 makes 1 mol of CO2, therefore 0.0450 mol of CuCO3 makes 0.0450 mol of CO2.Using the triangle, mass of CO2 = 44 x 0.0450 = 1.78 grams

CB
Answered by Clemmie B. Chemistry tutor

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