Question 2494a

May 23, 2017

You need to use 14.7 g of sodium.

Explanation:

There are four steps involved in this stoichiometry problem:

1. Write the balanced equation for the reaction.
2. Use the Ideal Gas Law to calculate the moles of ${\text{H}}_{2}$.
3. Use the molar ratio from the balanced equation to convert moles of ${\text{H}}_{2}$ to moles of $\text{Na}$.
4. Use the molar mass of $\text{Na}$ to convert moles of ${\text{Na}}_{3}$ to grams of $\text{Na}$.

Let's get started.

Step 1. Write the balanced chemical equation.

${\text{2Na" + 2"H"_2"O" → "2NaOH" + "H}}_{2}$

Step 2. Use the Ideal Gas Law to calculate the volume of ${\text{H}}_{2}$.

The Ideal Gas Law is

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

• $p$ = the pressure of the gas,
• $V$ = the volume of the gas,
• $n$ = the number of moles of the gas,
• $R$ = the universal gas constant
• $T$ = the temperature of the gas

We can rearrange the Ideal Gas Law to get

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

In this problem,

$p = \text{81.1 kPa}$
$V = \text{9.8 L}$
$R = \text{8.314 kPa·L·K"^"-1""mol"^"-1}$
$T = \text{(26.2 + 273.15) K = 299.35 K}$

n = (81.1 color(red)(cancel(color(black)("kPa"))) × 9.8 color(red)(cancel(color(black)("L"))))/(8.314 color(red)(cancel(color(black)("kPa"·"L"·"K"^"-1")))"mol"^"-1" × 299.35 color(red)(cancel(color(black)("K")))) = "0.319 mol"

Step 3. Convert moles of ${\text{H}}_{2}$ to moles of $\text{Na}$

The molar ratio of ${\text{Na:H}}_{2}$ is $2 : 1$.

$\text{Moles of Na" = 0.319 color(red)(cancel(color(black)("mol H"_2))) × ("2 mol Na")/(1 color(red)(cancel(color(black)("mol H"_2)))) = "0.639 mol Na}$

Step 4. Convert moles of $\text{Na}$ to grams of $\text{Na}$**

The molar mass of $\text{Na}$ is 22.99 g/mol.

0.639 color(red)(cancel(color(black)("mol Na"))) × ("22.99 g Na")/(1 color(red)(cancel(color(black)("mol Na")))) = "14.7 g Na"#

Here's a useful video on mass-volume conversions.