# Question 7ed6f

May 2, 2016

$\text{1.0 L}$ at STP

#### Explanation:

Start by writing the balanced chemical equation that describes this decomposition reaction

$2 {\text{HgO"_ ((s)) stackrel(color(white)(color(red)(Delta)aa))(->) 2"Hg"_ ((l)) + "O}}_{2 \left(g\right)}$

Notice that mercury(II) oxide decomposes to form liquid mercury and oxygen gas in a $2 : 1$ mole ratios. This means that when $2$ moles of mercury(II) oxide decompose, $1$ mole of oxygen gas is produced.

In other words, every mole of mercury(II) oxide that undergoes decomposition ends up producing $\frac{1}{2}$ moles of oxygen gas.

To determine how many moles of mercury)II) oxide react, use the compound's molar mass

20 color(red)(cancel(color(black)("g"))) * "1 mole HgO"/(216.6 color(red)(cancel(color(black)("g")))) = "0.0923 moles HgO"

Assuming that all the moles of mercury(II) oxide undergo decomposition, the reaction will produce

0.0923 color(red)(cancel(color(black)("moles HgO"))) * "1 mole O"_2/(2color(red)(cancel(color(black)("moles HgO")))) = "0.04615 moles O"_2#

Now, since you didn't provide the pressure and temperature at which the oxygen gas is collected, I'll assume that you're working at STP.

STP conditions are usually defined as a pressure of $\text{1 atm}$ and a temperature of ${0}^{\circ} \text{C}$. Under these conditions for pressure and temperature, one mole of any ideal gas occupies $\text{22.4 L}$ $\to$ this is known as the molar volume of a gas at STP.

In your case, the $0.0923$ moles of oxygen gas will occupy

$0.04615 \textcolor{red}{\cancel{\textcolor{b l a c k}{{\text{moles O"_2))) * overbrace("22.4 L"/(1color(red)(cancel(color(black)("mole O"_2)))))^(color(blue)("molar volume of a gas at STP")) = color(darkgreen)(ul(color(black)("1.0 L O}}_{2}}}}$

I'll leave the answer rounded to two sig figs, but keep in mind that you only have one significant figure for the mass of mercury(II) oxide.

SIDE NOTE It's worth noting that STP conditions are currently defined as a pressure of $\text{100 kPa}$ and a temperature of ${0}^{\circ} \text{C}$.

Under these conditions of pressure and temperature, one mole of any ideal gas occupies $\text{22.7 L}$.

However, more often than not, textbooks and online resources use the old definition of STP, at which one mole of any ideal gas occupies $\text{22.4 L}$.