Question

3 sur 32 dans la réaction suivante : caco₃ + 2 hcl → cacl₂ + co₂ + h₂o 25 g de carbonate de calcium (caco₃) sont dissous dans de l’acide chlorhydrique (hcl). si un volume de 500 ml de solution d’hcl est utilisé, quelle est la concentration molaire de la solution d’hcl ? 5 points a 0,25 mol/l c 1 mol/l e 3 mol/l b 0,5 mol/l d 1,5 mol/l

Explanation:

Step1: Calculate moles of \( \text{CaCO}_3 \)

Molar mass of \( \text{CaCO}_3 \) is \( 40 + 12 + 3\times16 = 100 \, \text{g/mol} \). Moles \( n = \frac{\text{mass}}{\text{molar mass}} = \frac{25 \, \text{g}}{100 \, \text{g/mol}} = 0.25 \, \text{mol} \).

Step2: Relate moles of \( \text{CaCO}_3 \) and \( \text{HCl} \)

From reaction \( \text{CaCO}_3 + 2 \text{HCl}
ightarrow \dots \), 1 mol \( \text{CaCO}_3 \) reacts with 2 mol \( \text{HCl} \). So moles of \( \text{HCl} = 2 \times 0.25 \, \text{mol} = 0.5 \, \text{mol} \).

Step3: Convert volume to liters

Volume of \( \text{HCl} \) solution is \( 500 \, \text{mL} = 0.5 \, \text{L} \).

Step4: Calculate molarity of \( \text{HCl} \)

Molarity \( M = \frac{\text{moles of solute}}{\text{volume of solution (L)}} = \frac{0.5 \, \text{mol}}{0.5 \, \text{L}} = 1 \, \text{mol/L} \)? Wait, no—wait, step2: 1 mol \( \text{CaCO}_3 \) needs 2 mol \( \text{HCl} \), so 0.25 mol \( \text{CaCO}_3 \) needs \( 2 \times 0.25 = 0.5 \) mol \( \text{HCl} \)? Wait, no, wait: 25g \( \text{CaCO}_3 \) is 0.25 mol. The stoichiometry is 1:2, so moles of HCl is \( 2 \times 0.25 = 0.5 \) mol? Wait, no, wait, 25g / 100g/mol = 0.25 mol \( \text{CaCO}_3 \). Then moles of HCl is 2 0.25 = 0.5 mol? Then volume is 0.5 L. So molarity is 0.5 mol / 0.5 L = 1 mol/L? Wait, no, wait, I think I made a mistake. Wait, 25g \( \text{CaCO}_3 \): molar mass 100g/mol, so 25/100 = 0.25 mol. Reaction: 1 mol \( \text{CaCO}_3 \) reacts with 2 mol HCl. So moles of HCl needed: 2 0.25 = 0.5 mol. Volume of HCl solution is 500 mL = 0.5 L. So molarity (M) = moles / volume (L) = 0.5 mol / 0.5 L = 1 mol/L? Wait, but let's check again. Wait, 25g \( \text{CaCO}_3 \) is 0.25 mol. 1 mol \( \text{CaCO}_3 \) reacts with 2 mol HCl, so 0.25 mol \( \text{CaCO}_3 \) reacts with 0.5 mol HCl. The volume of HCl is 0.5 L. So M = 0.5 mol / 0.5 L = 1 mol/L. So the correct answer is C. 1 mol/L. Wait, but let me recheck:

Molar mass of \( \text{CaCO}_3 \): Ca=40, C=12, O=16*3=48. Total 40+12+48=100 g/mol. Correct. 25g / 100g/mol = 0.25 mol. Correct. Stoichiometry: 1 mol \( \text{CaCO}_3 \) : 2 mol HCl. So 0.25 mol \( \text{CaCO}_3 \) : 0.5 mol HCl. Volume of HCl solution: 500 mL = 0.5 L. Molarity = 0.5 mol / 0.5 L = 1 mol/L. So answer is C. 1 mol/L.

Answer:

C. 1 mol/L