# Question e9d00

Apr 4, 2017

${\text{0.336 g NaHCO}}_{3}$

#### Explanation:

According to the balanced chemical equation that describes this decomposition reaction

$\textcolor{b l u e}{2} {\text{NaHCO"_ (3(s)) stackrel(color(white)(acolor(red)(Delta)aaa))(->) "Na"_ 2"CO"_ (3(s)) + "H"_ 2"O"_ ((l)) + "CO}}_{2 \left(g\right)}$ $\uparrow$

for every $\textcolor{b l u e}{2}$ moles of sodium hydrogen carbonate that undergo decomposition, you get $1$ mole of carbon dioxide.

Now, convert the mass of carbon dioxide to moles by using the molar mass of carbon dioxide

0.0880 color(red)(cancel(color(black)("g"))) * "1 mole CO"_2/(44.01 color(red)(cancel(color(black)("g")))) = "0.00200 moles CO"_2

You can say that in order for the reaction to produce $0.00200$ moles of carbon dioxide, it must consume

0.00200color(red)(cancel(color(black)("moles CO"_2))) * (color(blue)(2)color(white)(.)"moles NaHCO"_3)/(1color(red)(cancel(color(black)("mole CO"_2)))) = "0.00400 moles NaHCO"_3#

To convert this to grams, use the molar mass of sodium hydrogen carbonate

$0.00400 \textcolor{red}{\cancel{\textcolor{b l a c k}{\text{moles NaHCO"_3))) * "84.007 g"/(1color(red)(cancel(color(black)("mole NaHCO"_3)))) = color(darkgreen)(ul(color(black)("0.336 g}}}}$

The answer is rounded to three sig figs, the number of sig figs you have for the mass of carbon dioxide produced by the reaction.