Question #8bcb2

1 Answer
Mar 19, 2014

Answer:

The reaction releases 7.71 L of #"H"_2#.

Explanation:

Here are the steps to follow:

  1. First, write the balanced equation for the reaction.

  2. Use the molar mass to convert grams of #"Mg"# to moles of #"Mg"#.

  3. Use the molar ratio between #"Mg"# and #"H"_2# to get the moles of #"H"_2#.

  4. Finally, use the molar volume to convert moles of #"H"_2# to litres of #"H"_2#

.

Let’s see how this works.

1. The balanced equation is

#"Mg" +"2HCl" → "MgCl"_2 + "H"_2#

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

#8.25 color(red)(cancel(color(black)("g Mg"))) × "1 mol Mg"/(24.31 color(red)(cancel(color(black)("g Mg")))) = "0.339 mol Mg"#

3. Calculate the moles of #"H"_2#

The balanced equation tells us that #"1 mol Mg"# gives #"1 mol H"_2#. So,

#0.339 color(red)(cancel(color(black)("mol Mg"))) × ("1 mol H"_2)/(1 color(red)(cancel(color(black)("mol Mg")))) = "0.339 mol H"_2#

4. Calculate the volume of #"H"_2#

STP is defined as 0 °C and 1 bar.

Under these conditions, the volume of 1 mol of an ideal gas is 22.711 L.

To convert moles to litres at STP, we use the relation

#"22.711 L = 1 mol"#.

So,

#0.339 color(red)(cancel(color(black)("mol H"_2))) × ("22.711 L H"_2)/(1 color(red)(cancel(color(black)("mol H"_2)))) = "7.71 L H"_2#

Notice how we always write the conversion factors so that the units cancel to give the desired units for the answer.