Question

Question #4d229

Answer 1

`"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 `"18.0 g"` of carbon, `"C"`, were burned in the present of `"63.7 g"` of oxygen gas, `"O"_2`, and that `"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

`"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 carbon and oxygen gas react to form carbon dioxide, `"CO"_2`, then the mass of the product must be equal to `"66.0 g"`.

`"66.0 g of C and O"_2color(white)(.)"react " -> " 66.0 g of CO"_2color(white)(.)"are produced"`

`color(white)(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

`"C"_ ((s)) + "O"_ (2(g)) -> "CO"_ (2(g))`

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 `"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.