Question #c5367

There are four steps involved in this problem:

1. Write the balanced equation for the reaction.
2. Use the molar mass of #"Mg"# to calculate the moles of #"Mg"#.
3. Use the molar ratio of #"Mg":"H"_2# from the balanced equation to calculate the moles of #"H"_2#.
4. Use the Ideal Gas Law to calculate the volume of #"H"_2# at STP.

Step 1. Write the balanced chemical equation.

#"Mg" + "2HCl" → "MgCl"_2 + "H"_2#

Step 2. Calculate the moles of #"Mg"#.

#"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 #"H"_2#

#"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 #"H"_2#

The Ideal Gas Law is

#color(blue)(bar(ul(|color(white)(a/a)PV = nRTcolor(white)(a/a)|)))" "#

We can rearrange this equation to get

#V = (nRT)/P#

STP is defined as 1 bar and 0 °C, so

#n = "0.329 mol"#
#R = "0.083 14 bar·L·K"^"-1""mol"^"-1"#
#T = "0 °C" = "273.15 K"#
#P = "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"#