QuestionApril 21, 2025

How much CO_(2) (L) is produced when 2.10 kg of sodium bicarbonate reacts with excess hydrochloric acid at 25.0^circ C and 1.23 atm? 4.17times 10^1 4.17times 10^-1 4.98times 10^-1 4.98times 10^2 3.50

How much CO_(2) (L) is produced when 2.10 kg of sodium bicarbonate reacts with excess hydrochloric acid at 25.0^circ C and 1.23 atm? 4.17times 10^1 4.17times 10^-1 4.98times 10^-1 4.98times 10^2 3.50
How much CO_(2)
(L) is produced when 2.10 kg of sodium bicarbonate reacts with excess hydrochloric acid
at 25.0^circ C and 1.23 atm?
4.17times 10^1
4.17times 10^-1
4.98times 10^-1
4.98times 10^2
3.50

Solution
4.2(275 votes)

Answer

4.98 \times 10^{2} Explanation 1. Write the balanced chemical equation The reaction is 2 \text{NaHCO}_3 + 2 \text{HCl} \rightarrow 2 \text{NaCl} + 2 \text{H}_2\text{O} + 2 \text{CO}_2. 2. Calculate moles of sodium bicarbonate Molar mass of \text{NaHCO}_3 = 84.01 \, \text{g/mol}. Moles of \text{NaHCO}_3 = \frac{2100 \, \text{g}}{84.01 \, \text{g/mol}} = 25 \, \text{mol}. 3. Determine moles of CO_2 produced From the balanced equation, 2 moles of \text{NaHCO}_3 produce 2 moles of CO_2. Thus, 25 moles of \text{NaHCO}_3 produce 25 moles of CO_2. 4. Use ideal gas law to find volume of CO_2 **PV = nRT**. Solve for V: V = \frac{nRT}{P}. Given: n = 25 \, \text{mol}, R = 0.0821 \, \text{L atm/mol K}, T = 298 \, \text{K}, P = 1.23 \, \text{atm}. V = \frac{25 \times 0.0821 \times 298}{1.23} = 498.4 \, \text{L}.

Explanation

1. Write the balanced chemical equation<br /> The reaction is $2 \text{NaHCO}_3 + 2 \text{HCl} \rightarrow 2 \text{NaCl} + 2 \text{H}_2\text{O} + 2 \text{CO}_2$.<br /><br />2. Calculate moles of sodium bicarbonate<br /> Molar mass of $\text{NaHCO}_3 = 84.01 \, \text{g/mol}$. Moles of $\text{NaHCO}_3 = \frac{2100 \, \text{g}}{84.01 \, \text{g/mol}} = 25 \, \text{mol}$.<br /><br />3. Determine moles of $CO_2$ produced<br /> From the balanced equation, 2 moles of $\text{NaHCO}_3$ produce 2 moles of $CO_2$. Thus, 25 moles of $\text{NaHCO}_3$ produce 25 moles of $CO_2$.<br /><br />4. Use ideal gas law to find volume of $CO_2$<br /> **$PV = nRT$**. Solve for $V$: $V = \frac{nRT}{P}$. <br /> Given: $n = 25 \, \text{mol}$, $R = 0.0821 \, \text{L atm/mol K}$, $T = 298 \, \text{K}$, $P = 1.23 \, \text{atm}$.<br /> $V = \frac{25 \times 0.0821 \times 298}{1.23} = 498.4 \, \text{L}$.
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