# Question c5367

Feb 20, 2017

The volume of hydrogen would be 7.47 L.

#### Explanation:

There are four steps involved in this problem:

1. Write the balanced equation for the reaction.
2. Use the molar mass of $\text{Mg}$ to calculate the moles of $\text{Mg}$.
3. Use the molar ratio of ${\text{Mg":"H}}_{2}$ from the balanced equation to calculate the moles of ${\text{H}}_{2}$.
4. Use the Ideal Gas Law to calculate the volume of ${\text{H}}_{2}$ at STP.

Step 1. Write the balanced chemical equation.

${\text{Mg" + "2HCl" → "MgCl"_2 + "H}}_{2}$

Step 2. Calculate the moles of $\text{Mg}$.

$\text{Moles of Mg" = 8.0 color(red)(cancel(color(black)("g Mg"))) × "1 mol Mg"/(24.30 color(red)(cancel(color(black)("g Mg")))) = "0.329 mol Mg}$

Step 3. Calculate the moles of ${\text{H}}_{2}$

${\text{Moles of H"_2 = 0.329 color(red)(cancel(color(black)("mol Mg"))) × "1 mol H"_2/(1 color(red)(cancel(color(black)("mol Mg")))) = "0.329 mol H}}_{2}$

Step 4. Calculate the volume of ${\text{H}}_{2}$

The Ideal Gas Law is

$\textcolor{b l u e}{\overline{\underline{| \textcolor{w h i t e}{\frac{a}{a}} P V = n R T \textcolor{w h i t e}{\frac{a}{a}} |}}} \text{ }$

We can rearrange this equation to get

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

STP is defined as 1 bar and 0 °C, so

$n = \text{0.329 mol}$
$R = \text{0.083 14 bar·L·K"^"-1""mol"^"-1}$
$T = \text{0 °C" = "273.15 K}$
$P = \text{1 bar}$

V = (0.329 color(red)(cancel(color(black)("mol"))) × "0.083 14" color(red)(cancel(color(black)("bar")))·"L"·color(red)(cancel(color(black)("K"^"-1""mol"^"-1"))) × 273.15 color(red)(cancel(color(black)("K"))))/(1 color(red)(cancel(color(black)("bar")))) = "7.47 L"#