# When "11.6 g" of citric acid reacts with excess sodium bicarbonate, what mass of "CO"_2 is made in "g"? MW = "44 g/mol"

##### 1 Answer
Aug 14, 2017

You would make ${\text{8.0 g CO}}_{2}$, to two significant figures. The more exact value is $\text{7.975 g}$.

You should look for the keywords that identify what kind of problem this is.

Since it is mentioned that sodium bicarbonate is in excess, it means that the other reactant, citric acid, is the limiting reactant. That means it will be completely used up first, and that limits the amount of product you can make (${\text{CO}}_{2}$, $\text{H"_2"O}$, and ${\text{Na"_3"C"_6"H"_5"O}}_{7}$).

Your reaction was:

$3 {\text{NaHCO"_3(s) + "C"_6"H"_8"O"_7(aq) -> 3"CO"_2(g) + 3"H"_2"O"(l) + "Na"_3"C"_6"H"_5"O}}_{7} \left(a q\right)$

The remainder of the question is asking you to start with a known mass of ${\text{C"_6"H"_8"O}}_{7}$, citric acid, and use the concept of reaction stoichiometry to get to the mass of ${\text{CO}}_{2}$.

Note that the limiting reactant must be used when determining the theoretical yield.

Each coefficient can be considered the mols of the substance involved in the reaction, so for example...

• ${\text{3 mols NaHCO}}_{3} \left(s\right)$ react with ${\text{1 mol C"_6"H"_8"O}}_{7} \left(a q\right)$.
• ${\text{3 mols NaHCO}}_{3} \left(s\right)$ can produce ${\text{3 mols CO}}_{2} \left(g\right)$.
• ${\text{3 mols NaHCO}}_{3} \left(s\right)$ can produce $\text{3 mols H"_2"O} \left(l\right)$.

and so on. This allows you to write the following unit conversion (the same units cancel out) using the molar mass of citric acid:

$11.6 \cancel{{\text{g C"_6"H"_8"O"_7) xx overbrace(cancel("1 mol C"_6"H"_8"O"_7)/(192 cancel("g C"_6"H"_8"O"_7)))^"Molar mass of citric acid" xx ("3 mol CO"_2)/cancel("1 mol C"_6"H"_8"O}}_{7}}$

$= {\text{0.181 mols CO}}_{2} \left(g\right)$

That is how many mols of ${\text{CO}}_{2} \left(g\right)$ the reaction would produce, but you are asked for the mass in grams ($\text{g}$). So, you'll need to finish this by converting the mols of ${\text{CO}}_{2}$ to grams using the molar mass of ${\text{CO}}_{2}$.

0.181 cancel("mols CO"_2) xx overbrace("44 g CO"_2/cancel("1 mol CO"_2))^"Molar mass of carbon dioxide"

$= \underline{{\text{7.975 g CO}}_{2} \left(g\right)}$

From the $\text{44 g/mol}$, you only have two sig figs, so the mass of ${\text{CO}}_{2}$ produced is color(blue)("8.0 g" to two significant figures.