Question

10 g of (90 percent pure) \( \text{CaCO}_3 \), treated with excess of HCl, gives what mass of \( \text{CO}_2 \)?

• A. 4.4 g
• B. 5.6 g
• C. 3.6 g
• D. 6.4 g

Hint:

To find the mass of a product in a chemical reaction, first determine the moles of reactant used, use the mole ratio from the balanced equation, and then calculate the mass of the product using its molar mass.

Answer 1

The Correct Option is A

The reaction between \( \text{CaCO}_3 \) and \( \text{HCl} \) is as follows: \[ \text{CaCO}_3 + 2 \, \text{HCl} \rightarrow \text{CaCl}_2 + \text{CO}_2 + \text{H}_2\text{O} \] ### Step 1: Find the Molar Mass of \( \text{CaCO}_3 \) The molar mass of \( \text{CaCO}_3 \) is: \[ \text{Molar mass of } \text{CaCO}_3 = 40 + 12 + (3 \times 16) = 100 \, \text{g/mol} \] ### Step 2: Find the Moles of \( \text{CaCO}_3 \) Given that 10 g of \( \text{CaCO}_3 \) is used and it is 90% pure, the effective mass of \( \text{CaCO}_3 \) is: \[ \text{Mass of pure } \text{CaCO}_3 = 10 \times 0.90 = 9 \, \text{g} \] Now, calculate the moles of \( \text{CaCO}_3 \): \[ \text{Moles of } \text{CaCO}_3 = \frac{9}{100} = 0.09 \, \text{mol} \] ### Step 3: Moles of \( \text{CO}_2 \) From the balanced equation, 1 mole of \( \text{CaCO}_3 \) produces 1 mole of \( \text{CO}_2 \). Therefore, the moles of \( \text{CO}_2 \) produced are also 0.09 mol. ### Step 4: Find the Mass of \( \text{CO}_2 \) The molar mass of \( \text{CO}_2 \) is: \[ \text{Molar mass of } \text{CO}_2 = 12 + (2 \times 16) = 44 \, \text{g/mol} \] Thus, the mass of \( \text{CO}_2 \) is: \[ \text{Mass of } \text{CO}_2 = 0.09 \times 44 = 3.96 \, \text{g} \] So, the mass of \( \text{CO}_2 \) produced is approximately \( 4.4 \, \text{g} \). Thus, the correct answer is: \[ \boxed{(A) \, 4.4 \, \text{g}} \]