# Question 8eb7d

Jun 22, 2017

$45$ ${\text{g C"_2"H}}_{4}$

#### Explanation:

We're asked to find the mass (in $\text{g}$) of ${\text{C"_2"H}}_{4}$ required to react completely and form $140$ ${\text{g CO}}_{2}$.

To solve mass-mass stoichiometry problems like this one, we always use three steps:

1. Calculate the moles of the substance with known mass ${\text{CO}}_{2}$ using its molar mass.
2. Use the coefficients of the balanced chemical equation to find the relative number of moles of the substance in question.
3. Use the molar mass of that substance to find its mass, in $\text{g}$

1.

Let's first calculate the number of moles of carbon dioxide, using its molar mass ($44.01 \text{g"/"mol}$) and dimensional analysis:

140cancel("g Co"_2)((1color(white)(l)"mol CO"_2)/(44.01cancel("g CO"_2))) = color(red)(3.18 color(red)("mol CO"_2

2.

Now, using the coefficients of the equation, let's find the relative number of moles of the substance in question, ${\text{C"_2"H}}_{4}$:

color(red)(3.18)color(white)(l)cancel(color(red)("mol CO"_2))((1color(white)(l)"mol C"_2"H"_4)/(2cancel("mol CO"_2))) = color(green)(1.59 color(green)("mol C"_2"H"_4

3.

Finally, let's calculate the number of grams of ethylene, using its molar mass ($28.06 \text{g"/"mol}$):

color(green)(1.59)color(white)(l)cancel(color(green)("mol C"_2"H"_4))((28.06color(white)(l)"g C"_2"H"_4)/(1cancel("mol C"_2"H"_4))) = color(blue)(45 color(blue)("g C"_2"H"_4

rounded to two significant figures, the amount given in the problem (assuming the $0$ is not significant).

Thus, in order to produce $140$ grams of carbon dioxide according to this reaction, you would ideally need to burn color(blue)(45 sfcolor(blue)("grams of ethylene"#.