Question

9
Calcium carbonate reacts with dilute hydrochloric acid according to the equation shown
CaCO3 + 2HCl → CaCl2 + CO2 + H2O
10 g of calcium carbonate is reacted with 100 cm3 of 1 mol/dm3 hydrochloric acid.
The following statements are made.
1
1.2 dm3 of carbon dioxide is formed.
2
5.6 g of calcium chloride is formed.
3
4.8 g of carbon dioxide is formed.
4
No calcium carbonate is left when the reaction is completed.
Which statements about the reaction are correct?
А
1 and 2
B
1 and 4
с
2 and 3
D
3 and 4​

Answer 1

Answer:

(D) is the correct option

CaCO3 + 2HCl = CaCl2 + CO2 + H2O

no.of mole of HCl = 100×1/1000 = 0.1 mol

no.of mol of CaCO3 = 10/100 = 0.1 mol

CaCO3 is the limiting reactant.

Answer 2

(D) 3 and 4 - 4.8 g of carbon dioxide is formed and no calcium carbonate is left when the reaction is completed.

Explanation:

• $CaCO_{3} + 2HCl \rightarrow CaCl_{2} + CO_{2} + H_{2}O$

The molar mass of CaCO₃ = 40+12+3×16

• The molar mass of CaCO₃ = 100 g/mol

The molar mass of CaCl₂ = 40 + 2×35.5

• The molar mass of CaCl₂ = 111 g/ mol

For 1 mole, 100 g CaCO₃ produces 111 g of CaCl₂.

• Assuming for 10 g of CaCO₃, x g of CaCl₂ is produced.

Therefore,

• $x=111 \times \frac{10}{100}$
• $x = 11.1 g$

11.1 g of CaCl₂ is produced.,i.e., 2 statement is wrong.

• The molar mass of CO₂ = 12 + 16×2
• The molar mass of CO₂ = 44 g

For 1 mole, 100 g CaCO₃ produces 44 g of CO₂.

• Assuming for 10 g of CaCO₃, y g of CO₂ is produced.

Therefore,

• $y = 44 \times \frac{10}{100}$
• $y = 4.4 g$
• y ≈ 4.8 g

4.8 g of CO₂ is produced.,i.e., 1 statement is wrong and 3 statement is right.

Thus, the CaCO₃ compound is completely utilized in the reaction.,i.e., statement 4 is correct.