Question 4d229

Sep 22, 2016

$\text{66.0 g}$

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 $\text{18.0 g}$ of carbon, $\text{C}$, were burned in the present of $\text{63.7 g}$ of oxygen gas, ${\text{O}}_{2}$, and that $\text{15.7 g}$ of oxygen gas remained unreacted, i.e. they did not take part in the reaction.

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

$\text{18.0 g " + " 48.0 g" = "66.0 g } \to$ 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 $\text{66.0 g}$ of carbon and oxygen gas react to form carbon dioxide, ${\text{CO}}_{2}$, then the mass of the product must be equal to $\text{66.0 g}$.

$\text{66.0 g of C and O"_2color(white)(.)"react " -> " 66.0 g of CO"_2color(white)(.)"are produced}$

$\textcolor{w h i t e}{a}$
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

${\text{C"_ ((s)) + "O"_ (2(g)) -> "CO}}_{2 \left(g\right)}$

Notice that $1$ mole of carbon reacts with $1$ mole of oxygen gas to produce $1$ mole of carbon dioxide.

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 $1 : 1$ mole ratios that exist across the board tell you that the reaction will produce $1.50$ moles of carbon dioxide.

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 $\text{66.0 g}$ of carbon dioxide, but this time we had a bit more work to do to find the answer.

As a conclusion, always keep an eye out for cases where the answer can be found by using the Law of Mass Conservation.