A student reacts 10.0 g of CaCO_3 with excess HCl to produce CaCl_2, CO_2, and H_2O. If the balanced equation is CaCO_3 + 2 HCl → CaCl_2 + CO_2 + H_2O, what is the theoretical yield of CaCl_2 (in grams)? (Molar masses: CaCO_3 = 100.1 g/mol, CaCl_2 = 110.98 g/mol)

• A. 22.2 g
• B. 10.0 g
• C. 5.5 g
• D. 11.1 g

Answer: (D)

Verified step by step guidance

1. Identify the given information: mass of CaCO_3 = 10.0 g, molar mass of CaCO_3 = 100.1 g/mol, molar mass of CaCl_2 = 110.98 g/mol, and the balanced chemical equation: CaCO_3 + 2 HCl → CaCl_2 + CO_2 + H_2O.
2. Calculate the number of moles of CaCO_3 reacted using the formula: \(\text{moles of CaCO}_3 = \frac{\text{mass of CaCO}_3}{\text{molar mass of CaCO}_3} = \frac{10.0\ \text{g}}{100.1\ \text{g/mol}}\).
3. Use the stoichiometric relationship from the balanced equation to find the moles of CaCl_2 produced. Since the ratio of CaCO_3 to CaCl_2 is 1:1, the moles of CaCl_2 formed will be equal to the moles of CaCO_3 reacted.
4. Calculate the theoretical mass of CaCl_2 produced by multiplying the moles of CaCl_2 by its molar mass: \(\text{mass of CaCl}_2 = \text{moles of CaCl}_2 \times 110.98\ \text{g/mol}\).
5. The result from step 4 gives the theoretical yield of CaCl_2 in grams, which can be compared to the given answer choices.