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QUESTION

# Mg metal reacts with HCI to produce hydrogen gas. Mg (s) + 2HCl (aq) ---&gt; MgCl2 (aq) + H2 (g) What volume,in liters, of H2 at STP is released when 8.25g of Mg reacts?

The reaction releases 7.71 L of "H"_2.

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.