# Question 461b0

Feb 11, 2018

$\text{0.31 g}$

#### Explanation:

The idea here is that under STP conditions, which are defined as a pressure of $\text{100 kPa}$ and a temperature of ${0}^{\circ} \text{C}$, $1$ mole of any ideal gas occupies ${\text{22.71 dm}}^{3}$ $\to$ this is known as the molar volume of a gas at STP .

Since you know that

${\text{1 dm"^3 = 10^3 quad "cm}}^{3}$

you can say that at STP, $1$ mole of any ideal gas occupies $2.271 \cdot {10}^{4}$ ${\text{cm}}^{3}$.

You can use the molar volume of a gas at STP to find the number of moles of oxygen as present in your sample.

220 color(red)(cancel(color(black)("cm"^3))) * "1 mole O"_2/(2.271 * 10^4 color(red)(cancel(color(black)("cm"^3)))) = 9.687 * 10^(-3) quad "moles O"_2#

To convert the number of moles of oxygen gas to moles, you can use the molar mass of oxygen gas.

$9.687 \cdot {10}^{- 3} \textcolor{red}{\cancel{\textcolor{b l a c k}{\text{moles O"_2))) * "32.0 g"/(1color(red)(cancel(color(black)("mole O"_2)))) = color(darkgreen)(ul(color(black)("0.31 g}}}}$

The answer is rounded to two sig figs, the number of sig figs you have for the volume of oxygen gas.