Question

15 g of $ \text{CaCO}_3 $ completely reacts with

• A. 6.95 g of HCl
• B. 10.95 g of HCl
• C. 11.95 g of HCl
• D. 1.15 g of HCl

Hint:

To calculate the amount of a reactant needed, use the molar ratio from the balanced chemical equation along with the molar mass of the substances involved.

Answer 1

The Correct Option is B

To determine the amount of HCl that reacts completely with 15 g of CaCO₃, we can use the stoichiometry of the reaction:

CaCO₃ + 2HCl → CaCl₂ + CO₂ + H₂O

The first step is to calculate the molar mass of CaCO₃ and HCl:

• Molar mass of CaCO₃ = 40 (Ca) + 12 (C) + 3×16 (O) = 100 g/mol
• Molar mass of HCl = 1 (H) + 35.5 (Cl) = 36.5 g/mol

Next, find the moles of CaCO₃ in 15 g:

Moles of CaCO₃ = 15 g100 g/mol = 0.15 moles

According to the balanced equation, 1 mole of CaCO₃ reacts with 2 moles of HCl. Thus, 0.15 moles of CaCO₃ will require:

0.15 moles of CaCO₃ × 2 moles of HCl/mole of CaCO₃ = 0.30 moles of HCl

Now, convert moles of HCl to grams:

Mass of HCl = 0.30 moles × 36.5 g/mol = 10.95 g

Therefore, 15 g of CaCO₃ completely reacts with 10.95 g of HCl.

Answer 2

The reaction between calcium carbonate (\( \text{CaCO}_3 \)) and hydrochloric acid (HCl) is given by the following equation: \[ \text{CaCO}_3 + 2 \, \text{HCl} \rightarrow \text{CaCl}_2 + \text{H}_2\text{O} + \text{CO}_2 \] 
From the balanced equation, it is clear that 1 mole of \( \text{CaCO}_3 \) reacts with 2 moles of HCl. 
Now, let's calculate the moles of \( \text{CaCO}_3 \) in 15 g: 
Molar mass of \( \text{CaCO}_3 \) = 40 + 12 + (3 \times 16) = 100 g/mol. 
Moles of \( \text{CaCO}_3 \) = \( \frac{15 \, \text{g}}{100 \, \text{g/mol}} = 0.15 \, \text{mol} \). 
From the equation, 1 mole of \( \text{CaCO}_3 \) reacts with 2 moles of HCl, so 0.15 moles of \( \text{CaCO}_3 \) will react with: \( 0.15 \, \text{mol} \times 2 = 0.30 \, \text{mol} \) of HCl. Now, calculate the mass of 0.30 moles of HCl: 
Molar mass of HCl = 36.5 g/mol. Mass of HCl = \( 0.30 \, \text{mol} \times 36.5 \, \text{g/mol} = 10.95 \, \text{g} \). 
Thus, the mass of HCl required is 10.95 g.