Question

Which of the following statement(s) is/are correct for decomposition of $NaHCO_3$?

• A. 84 g of $NaHCO_3$ gives 106 g of $Na_2CO_3$
• B. 16.8g of $NaHCO_3$ gives 0.1 mole of $Na_2CO_3$
• C. 0.1 mole of $NaHCO_3$ gives 1.12L of $CO_2$ at STP
• D. 84 g of $NaHCO_3$ gives 22.4L of $CO_2$ at STP

Answer 1

Answer: (B), (C)

Statements (b) and (c) are correct.

Answer 2

The decomposition of sodium bicarbonate ($NaHCO_3$) follows the balanced chemical equation: $2NaHCO_3(s) \rightarrow Na_2CO_3(s) + H_2O(g) + CO_2(g)$

Molar mass of $NaHCO_3 = 23 (\text{Na}) + 1 (\text{H}) + 12 (\text{C}) + 3 \times 16 (\text{O}) = 84$ g/mol. Molar mass of $Na_2CO_3 = 2 \times 23 (\text{Na}) + 12 (\text{C}) + 3 \times 16 (\text{O}) = 106$ g/mol. Molar volume of gas at STP = 22.4 L/mol.

(a) 84 g of $NaHCO_3$ gives 106 g of $Na_2CO_3$ 84 g of $NaHCO_3$ is 1 mole. From the equation, 2 moles $NaHCO_3$ produce 1 mole $Na_2CO_3$. So, 1 mole $NaHCO_3$ produces 0.5 mole $Na_2CO_3$. Mass of 0.5 mole $Na_2CO_3 = 0.5 \text{ mol} \times 106 \text{ g/mol} = 53$ g. This statement is incorrect.

(b) 16.8g of $NaHCO_3$ gives 0.1 mole of $Na_2CO_3$ 16.8 g of $NaHCO_3 = 16.8 \text{ g} / 84 \text{ g/mol} = 0.2$ moles. 0.2 moles $NaHCO_3$ produce $(0.2 / 2) = 0.1$ mole of $Na_2CO_3$. This statement is correct.

(c) 0.1 mole of $NaHCO_3$ gives 1.12L of $CO_2$ at STP 0.1 mole $NaHCO_3$ produce $(0.1 / 2) = 0.05$ mole of $CO_2$. Volume of $CO_2$ at STP $= 0.05 \text{ mol} \times 22.4 \text{ L/mol} = 1.12$ L. This statement is correct.

(d) 84 g of $NaHCO_3$ gives 22.4L of $CO_2$ at STP 84 g of $NaHCO_3$ is 1 mole. 1 mole $NaHCO_3$ produces $(1 / 2) = 0.5$ mole of $CO_2$. Volume of $CO_2$ at STP $= 0.5 \text{ mol} \times 22.4 \text{ L/mol} = 11.2$ L. This statement is incorrect.