Question 8bcb2

Mar 19, 2014

The reaction releases 7.71 L of ${\text{H}}_{2}$.

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 $\text{Mg}$ to moles of $\text{Mg}$.

3. Use the molar ratio between $\text{Mg}$ and ${\text{H}}_{2}$ to get the moles of ${\text{H}}_{2}$.

4. Finally, use the molar volume to convert moles of ${\text{H}}_{2}$ to litres of ${\text{H}}_{2}$

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Let’s see how this works.

1. The balanced equation is

${\text{Mg" +"2HCl" → "MgCl"_2 + "H}}_{2}$

2. Calculate the moles of $\text{Mg}$.

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 ${\text{H}}_{2}$

The balanced equation tells us that $\text{1 mol Mg}$ gives ${\text{1 mol H}}_{2}$. 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 ${\text{H}}_{2}$

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

$\text{22.711 L = 1 mol}$.

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.