Problemas
10. This chemical reaction follows the law of conservation of mass. How much CaCO_(3) will be formed if 2.2 grams of CaO are used CaO(s)+CO_(2)(g)CaCO_(3)(s) for this reaction? Assume there are enough reactants to complete the reaction. 1.1 grams 2.2 grams 3.9 grams 4.3 grams
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To solve this problem, we need to use the concept of stoichiometry and the balanced chemical equation.Given information:- The balanced chemical equation is: CaO(s) + CO2(g) → CaCO3(s)- 2.2 grams of CaO are used.Step 1: Calculate the molar mass of CaO and CaCO3.Molar mass of CaO = 56.08 g/mol (Ca) + 16.00 g/mol (O) = 72.08 g/molMolar mass of CaCO3 = 56.08 g/mol (Ca) + 12.01 g/mol (C) + 3 × 16.00 g/mol (O) = 100.09 g/molStep 2: Calculate the number of moles of CaO.Number of moles of CaO = 2.2 g / 72.08 g/mol = 0.0306 molStep 3: Use the stoichiometry of the balanced equation to calculate the number of moles of CaCO3 formed.According to the balanced equation, 1 mol of CaO produces 1 mol of CaCO3.Number of moles of CaCO3 formed = 0.0306 molStep 4: Calculate the mass of CaCO3 formed.Mass of CaCO3 formed = Number of moles of CaCO3 × Molar mass of CaCO3Mass of CaCO3 formed = 0.0306 mol × 100.09 g/mol = 3.1 gTherefore, the correct answer is 3.1 grams.