# Question 12c2c

Jul 8, 2017

$\text{26 g}$

#### Explanation:

The idea here is that you're looking for the mass of the metal that will displace exactly $1.008$ parts by mass of elemental hydrogen, $\text{H}$.

The first thing that you need to do here is to use the ideal gas law equation

$\textcolor{b l u e}{\underline{\textcolor{b l a c k}{P V = n R T}}}$

Here

• $P$ is the pressure of the gas
• $V$ is the volume it occupies
• $n$ is the number of moles of gas present in the sample
• $R$ is the universal gas constant, equal to $0.0821 \left(\text{atm L")/("mol K}\right)$
• $T$ is the absolute temperature of the gas

to find the number of moles of hydrogen gas, ${\text{H}}_{2}$, present in your sample under normal conditions for pressure and temperature, i.e. at NTP, which are defined as a pressure of $\text{1 atm}$ and a temperature of ${20}^{\circ} \text{C = 293.15 K}$,

Rearrange the equation to solve for $n$

$P V = n R T \implies n = \frac{P V}{R T}$

Plug in your values to find

n = (1 color(red)(cancel(color(black)("atm"))) * 1.12color(red)(cancel(color(black)("L"))))/(0.0821(color(red)(cancel(color(black)("atm"))) * color(red)(cancel(color(black)("L"))))/("mol" * color(red)(cancel(color(black)("K")))) * 293.15 color(red)(cancel(color(black)("K")))) = "0.04654 moles H"_2

The number of moles of elemental hydrogen present in your sample will be equal to

0.04654 color(red)(cancel(color(black)("moles H"_2))) * "2 moles H"/(1color(red)(cancel(color(black)("mole H"_2)))) ~~ "0.09308 moles H"

Convert this to grams by using the molar mass of elemental hydrogen

0.09308 color(red)(cancel(color(black)("moles H"))) * "1.008 g"/(1color(red)(cancel(color(black)("mole H")))) = "0.09382 g"#

Now, if $\text{0.09382 g}$ of hydrogen were displaced by $\text{2.4 g}$ of metal, it follows that $\text{1.008 g}$ of hydrogen will be displaced by

$1.008 \textcolor{red}{\cancel{\textcolor{b l a c k}{\text{g H"))) * "2.4 g metal"/(0.09382color(red)(cancel(color(black)("g H")))) = color(darkgreen)(ul(color(black)("26 g metal}}}}$

The answer is rounded to two sig figs, the number of significant figures you have for the mass of the metal.

$\textcolor{w h i t e}{a}$
SIDE NOTE I suspect that the problem was designed with the old STP conditions in mind, i.e. a pressure of $\text{1 atm}$ and a temperature of ${0}^{\circ} \text{C}$.

Under these conditions for pressure and temperature, the molar volume of an ideal gas is equal to $\text{22.4 L}$.

In this case, the number of moles of elemental hydrogen would come out to be equal to $0.100$, which would produce an equivalent mass of $\text{24 g}$ for the metal.