Question

Question #2494a

Answer 1

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 `"H"_2`.
3. Use the molar ratio from the balanced equation to convert moles of `"H"_2` to moles of `"Na"`.
4. Use the molar mass of `"Na"` to convert moles of `"Na"_3` to grams of `"Na"`.

Let's get started.

Step 1. Write the balanced chemical equation.

`"2Na" + 2"H"_2"O" → "2NaOH" + "H"_2`

Step 2. Use the Ideal Gas Law to calculate the volume of `"H"_2`.

The Ideal Gas Law is

`color(blue)(bar(ul(|color(white)(a/a)pV = nRTcolor(white)(a/a)|)))" "`

• `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

`color(blue)(bar(ul(|color(white)(a/a)n = (pV)/(RT)color(white)(a/a)|)))" "`

In this problem,

`p = "81.1 kPa"`
`V = "9.8 L"`
`R = "8.314 kPa·L·K"^"-1""mol"^"-1"`
`T = "(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 `"H"_2` to moles of `"Na"`

The molar ratio of `"Na:H"_2` is `2:1`.

`"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 `"Na"` to grams of `"Na"`**

The molar mass of `"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.