Question

Elixirs such as Alka-Seltzer use the reaction of sodium bicarbonate with citric acid in aqueoussolution to produce a fizz:

$$\begin{gathered} \mathbf{3}{\mathbf{\text{NaHCO}}}_{\mathbf{3}}\mathbf{\left(}\mathbf{a}\mathbf{q}\mathbf{\right)}\mathbf{+}{\mathbf{\text{C}}}_{\mathbf{6}}{\mathbf{\text{H}}}_{\mathbf{8}}{\mathbf{\text{O}}}_{\mathbf{7}}\mathbf{\left(}\mathbf{a}\mathbf{q}\mathbf{\right)}\mathbf{\to }\mathbf{3}{\mathbf{\text{CO}}}_{\mathbf{2}}\mathbf{\left(}\mathbf{g}\mathbf{\right)}\mathbf{+}\mathbf{3}{\mathbf{\text{H}}}_{\mathbf{2}}\mathbf{\text{O}}\mathbf{\left(}\mathbf{l}\mathbf{\right)}\mathbf{+} \\ {\mathbf{\text{Na}}}_{\mathbf{3}}{\mathbf{\text{C}}}_{\mathbf{6}}{\mathbf{\text{H}}}_{\mathbf{5}}{\mathbf{\text{O}}}_{\mathbf{7}}\mathbf{\left(}\mathbf{a}\mathbf{q}\mathbf{\right)} \end{gathered}$$

a. What mass of${\mathbf{\text{C}}}_{\mathbf{6}}{\mathbf{\text{H}}}_{\mathbf{8}}{\mathbf{\text{O}}}_{\mathbf{7}}$should be used for$\mathbf{1}\mathbf{.}\mathbf{0}\mathit{x}{\mathbf{10}}^{\mathbf{2}}{\mathbf{\text{mgNaHCO}}}_{\mathbf{3}}$ every?
b. What mass of${\mathbf{\text{CO}}}_{\mathbf{2}}\left(\text{g}\right)$ could be produced from such a mixture?

Short answer

1. The mass of${\mathbf{\text{C}}}_{\mathbf{6}}{\mathbf{\text{H}}}_{\mathbf{8}}{\mathbf{\text{O}}}_{\mathbf{7}}$should be used for every$1.0\mathrm{x}{10}^2\mathrm{mg}{\mathrm{NaHCO}}_3$is$\mathbf{76}\mathbf{}{\mathbf{\text{mgC}}}_{\mathbf{6}}{\mathbf{\text{H}}}_{\mathbf{8}}{\mathbf{\text{O}}}_{\mathbf{7}}$.
2. The mass of $\mathit{C}{\mathit{O}}_{\mathbf{2}}\left(g\right)$could be produced from such a mixture is$\mathbf{52}{\mathbf{\text{mgCO}}}_{\mathbf{2}}$ .

Step-by-step solution

Definition

The mass of any material expressed in atomic mass units is quantitatively equivalent to its molar mass in grams per mole.

Determination of Molar mass of C6H8O7and NaHCO3

a)

Molar mass of:

$$\begin{gathered} =12.01\raisebox{1ex}{$\text{g}$}\!\left/ \!\raisebox{-1ex}{$\text{mol}$}\right.\times 6+1.01\raisebox{1ex}{$\text{g}$}\!\left/ \!\raisebox{-1ex}{$\text{mol}$}\right.\times 8+16.00\raisebox{1ex}{$\text{g}$}\!\left/ \!\raisebox{-1ex}{$\text{mol}$}\right.\times 7 \\ =192.14\raisebox{1ex}{$\text{g}$}\!\left/ \!\raisebox{-1ex}{$\text{mol}$}\right. \end{gathered}$$

Thus the molar mass of ${\mathbf{\text{C}}}_{\mathbf{6}}{\mathbf{\text{H}}}_{\mathbf{8}}{\mathbf{\text{O}}}_{\mathbf{7}}$ is$\mathbf{=}\mathbf{192}\mathbf{.}\mathbf{14}\mathbf{}\raisebox{1ex}{$\mathbf{\text{g}}$}\!\left/ \!\raisebox{-1ex}{$\mathbf{\text{mol}}$}\right.$.

Molar mass of ${\mathbf{\text{NaHCO}}}_{\mathbf{3}}$:

$$\begin{gathered} =23.0\raisebox{1ex}{$\text{g}$}\!\left/ \!\raisebox{-1ex}{$\text{mol}$}\right.+1.01\raisebox{1ex}{$\text{g}$}\!\left/ \!\raisebox{-1ex}{$\text{mol}$}\right.+12.01\raisebox{1ex}{$\text{g}$}\!\left/ \!\raisebox{-1ex}{$\text{mol}$}\right.+16.00\raisebox{1ex}{$\text{g}$}\!\left/ \!\raisebox{-1ex}{$\text{mol}$}\right.\times 3 \\ =84.02\raisebox{1ex}{$\text{g}$}\!\left/ \!\raisebox{-1ex}{$\text{mol}$}\right. \end{gathered}$$

Thus the molar mass of ${\mathbf{\text{NaHCO}}}_{\mathbf{3}}$ is $\mathbf{84}\mathbf{.}\mathbf{02}\mathbf{}\raisebox{1ex}{$\mathbf{\text{g}}$}\!\left/ \!\raisebox{-1ex}{$\mathbf{\text{mol}}$}\right.$.

Calculation of Mass of C6H8O7

We already know that${\mathbf{\text{NaHCO}}}_{\mathbf{3}}$ is3 times the molof ${\mathbf{\text{C}}}_{\mathbf{6}}{\mathbf{\text{H}}}_{\mathbf{8}}{\mathbf{\text{O}}}_{\mathbf{7}}$

role="math" localid="1648724028337" $$\begin{gathered} 0.1g\times {\mathrm{NaHCO}}_3\times \frac{1\mathrm{mol}{\mathrm{NaHCO}}_3}{84.02g}\times \frac{1\mathrm{mol}{C}_6{H}_8{O}_7}{3\mathrm{mol}{\mathrm{NaHCO}}_3}\times \frac{192.14g}{1mol} \\ =0.076g \end{gathered}$$

$=76\text{mg}{\mathrm{C}}_6{\mathrm{H}}_8{\mathrm{O}}_7$

Therefore the mass of ${\mathbf{\text{C}}}_{\mathbf{6}}{\mathbf{\text{H}}}_{\mathbf{8}}{\mathbf{\text{O}}}_{\mathbf{7}}$is role="math" localid="1648724247756" $\mathbf{76}{\mathbf{\text{mgC}}}_{\mathbf{6}}{\mathbf{\text{H}}}_{\mathbf{8}}{\mathbf{\text{O}}}_{\mathbf{7}}$

Determination of Mass of CO2

b)

Molar mass of${\text{CO}}_2$:

$\text{12.01}\raisebox{1ex}{$\text{g}$}\!\left/ \!\raisebox{-1ex}{$\text{mol}$}\right.\text{+16.00}\raisebox{1ex}{$\text{g}$}\!\left/ \!\raisebox{-1ex}{$\text{mol}$}\right.\text{×2=44.01}\raisebox{1ex}{$\text{g}$}\!\left/ \!\raisebox{-1ex}{$\text{mol}$}\right.$

We already know that ${\text{NaHCO}}_3$is the same as mol of ${\text{CO}}_2$

role="math" localid="1648723572408" $$\begin{gathered} 0.1g\times {\mathrm{NaHCO}}_3\times \frac{1\mathrm{mol}{\mathrm{NaHCO}}_3}{84.02g}\times \frac{3\mathrm{mol}C{O}_2}{3\mathrm{mol}{\mathrm{NaHCO}}_3}\times \frac{44.01g}{1mol} \\ =0.052g \\ =52{\text{mgCO}}_2 \end{gathered}$$

Therefore, the mass role="math" localid="1648723649919" ${\mathbf{\text{CO}}}_{\mathbf{2}}\left(g\right)$is${\mathbf{\text{52mgCO}}}_{\mathbf{2}}$ .