Question

Question #8bcb2

Answer 1

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.