Question #e9d00

According to the balanced chemical equation that describes this decomposition reaction

#color(blue)(2)"NaHCO"_ (3(s)) stackrel(color(white)(acolor(red)(Delta)aaa))(->) "Na"_ 2"CO"_ (3(s)) + "H"_ 2"O"_ ((l)) + "CO"_ (2(g))# #uarr#

for every #color(blue)(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 color(red)(cancel(color(black)("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.