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

To solve this problem, we need to use the concept of stoichiometry and the balanced chemical equation.<br /><br />Given information:<br />- The balanced chemical equation is: CaO(s) + CO2(g) → CaCO3(s)<br />- 2.2 grams of CaO are used.<br /><br />Step 1: Calculate the molar mass of CaO and CaCO3.<br />Molar mass of CaO = 56.08 g/mol (Ca) + 16.00 g/mol (O) = 72.08 g/mol<br />Molar mass of CaCO3 = 56.08 g/mol (Ca) + 12.01 g/mol (C) + 3 × 16.00 g/mol (O) = 100.09 g/mol<br /><br />Step 2: Calculate the number of moles of CaO.<br />Number of moles of CaO = 2.2 g / 72.08 g/mol = 0.0306 mol<br /><br />Step 3: Use the stoichiometry of the balanced equation to calculate the number of moles of CaCO3 formed.<br />According to the balanced equation, 1 mol of CaO produces 1 mol of CaCO3.<br />Number of moles of CaCO3 formed = 0.0306 mol<br /><br />Step 4: Calculate the mass of CaCO3 formed.<br />Mass of CaCO3 formed = Number of moles of CaCO3 × Molar mass of CaCO3<br />Mass of CaCO3 formed = 0.0306 mol × 100.09 g/mol = 3.1 g<br /><br />Therefore, the correct answer is 3.1 grams.