# How much volume would 63.9 g of carbon monoxide take up?

Jan 28, 2017

The volume of $\text{63.9 g}$ carbon dioxide at STP is $\text{32.6 L}$.

#### Explanation:

Temperature affects volume. Since no temperature was given, I'm going to use STP.

The density of carbon dioxide can be used to answer this question.

We have been given the mass of $\text{CO"_2}$ gas. We can look up the density at STP, which is $\text{1.96 g/L}$. http://nobel.scas.bcit.ca/chem0010/unit7/mole8.htm

We now have density and mass.

The Density Formula

$\text{density"="mass"/"volume}$

To determine volume, rearrange the formula to solve for volume, then substitute the known values and solve.

$\text{volume"="mass"/"density}$

$\text{volume"=(63.9color(white)(.)cancel"g")/(1.96cancel"g"/"L")="32.6 L}$ rounded to three significant figures

Jan 28, 2017

51.1 liters at STP

#### Explanation:

The molar mass of Carbon Monoxide $C O$ is 28 grams/mole

1 x C = 1 x 12 = 12
1 x O = 1 x 16 = 16
Total = 28 grams/ mole

The number of moles is found by dividing the mass by the grams/mole

$\frac{63.9}{28} = 2.28$ moles.

At STP ( Standard Temperature Pressure) The molar volume is 22.4 liters/mole

to find the volume multiple the number of moles time the molar volume.

# 2.28 xx 22.4 = 51.1 liters.