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0.5 moles of gas A and $x$ moles of gas B together exert a pressure of 200 Pa in a 10 m$^3$ container at 1000 K. Given that $R$ is the gas constant in J K$^{-1}$ mol$^{-1}$, the value of $x$ is:
Apply the ideal gas law: $PV = n_{\text{total}}RT$
$n_{\text{total}} = 0.5 + x$
$200 \times 10 = (0.5 + x) \times R \times 1000$
$2000 = 1000R(0.5 + x)$
$2 = R(0.5 + x) = 0.5R + Rx$
$Rx = 2 - 0.5R$
$x = \dfrac{2 - 0.5R}{R} = \dfrac{4 - R}{2R}$
Moles of CO$_2 = \dfrac{0.25 \times 10^{-3}\ \text{L}}{25.0\ \text{L mol}^{-1}} = 1.0 \times 10^{-5}\ \text{mol}$
Reaction: $\text{NaHCO}_3 + \text{H}^+ \to \text{Na}^+ + \text{H}_2\text{O} + \text{CO}_2$
Moles of NaHCO$_3 = 1.0 \times 10^{-5}\ \text{mol}$
Mass of NaHCO$_3 = 1.0 \times 10^{-5} \times 84 = 8.4 \times 10^{-4}\ \text{g} = 0.84\ \text{mg}$
$\% = \dfrac{0.84}{10} \times 100 = \mathbf{8.4\%}$
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