Question #4d229
1 Answer
Explanation:
All you have to do here is use the Law of Mass Conservation to help you find the mass of carbon dioxide produced by your reaction.
You are told that
You can say that the mass of oxygen gas that took part in the reaction was
#overbrace("63.7 g")^(color(blue)("total mass of O"_2)) - overbrace("15.7 g")^(color(purple)("remained unreacted")) = "48.0 g O"_2#
Now, the total mass of carbon and oxygen gas that took part in the reaction was
#"18.0 g " + " 48.0 g" = "66.0 g " -># the mass of the reactants
As you know, mass is conserved in a chemical equation as described by the aforementioned Law of Mass Conservation.
This basically means that if
#"66.0 g of C and O"_2color(white)(.)"react " -> " 66.0 g of CO"_2color(white)(.)"are produced"#
LET'S PROVE THIS BY USING MOLES
One way to prove that this is the answer is by using the balanced chemical equation that describes this synthesis reaction
#"C"_ ((s)) + "O"_ (2(g)) -> "CO"_ (2(g))#
Notice that
Use the molar masses of the two reactants to convert the grams to moles -- keep in mind that you only use the mass of oxygen that takes part in the reaction here, which does not include what remains unreacted!
#18.0 color(red)(cancel(color(black)("g"))) * "1 mole C"/(12.0 color(red)(cancel(color(black)("g")))) = "1.50 moles C"#
#48.0 color(red)(cancel(color(black)("g"))) * "1 mole O"_2/(32.0 color(red)(cancel(color(black)("g")))) = "1.50 moles O"_2#
As you can see, the two reactants will be completely consumed by the reaction. Consequently, the
To convert this to grams, use the molar mass of carbon dioxide
#1.50 color(red)(cancel(color(black)("moles CO"_2))) * "44.0 g"/(1color(red)(cancel(color(black)("mole CO"_2)))) = "66.0 g"#
Once again, we get that the reaction produced
As a conclusion, always keep an eye out for cases where the answer can be found by using the Law of Mass Conservation.