Question #2494a

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.