# Question 36e9a

Nov 3, 2015

$\text{1.025 L}$

#### Explanation:

Start by writing a balanced chemical equation for this single replacement reaction

${\text{Mg"_text((s]) + 2"HCl"_text((aq]) -> "MgCl"_text(2(aq]) + "H}}_{\textrm{2 \left(g\right]}} \uparrow$

Notice that you have a $1 : 1$ mole ratio between
magnesium and hydrogen gas. This means that the reaction will produce $1$ mole of hdyrogen gas for every $1$ mole of magnesium that takes part in the reaction.

Use magnesium's molar mass to determine how many moles you have in that sample

1.112color(red)(cancel(color(black)("g"))) * "1 mole Mg"/(24.305color(red)(cancel(color(black)("g")))) = "0.04575 moles Mg"

The aforementioned mole ratio tells you tha tthe reaction will produce

0.04575color(red)(cancel(color(black)("moles Mg"))) * "1 mole H"_2/(1color(red)(cancel(color(black)("mole Mg")))) = "0.04575 moles H"_2

Now, the conditions for pressure and temperature given to you correspond to the old STP conditions at which one mole of any ideal gas occupied exactly $\text{22.4 L}$.

STP conditions have been changed to $\text{100 kPa}$ and ${0}^{\circ} \text{C}$, but that doesn't change the fact that at $\text{1 atm}$ and ${0}^{\circ} \text{C}$, one mole of any ideal gas occupies $\text{22.4 L}$.

This means that the reaction will produce a volume of

0.04575color(red)(cancel(color(black)("moles"))) * "22.4 L"/(1color(red)(cancel(color(black)("mole")))) = "1.0248 L"#

I'll leave the answer rounded to four sig figs, the number of sig figs you have for the mass of magnesium

${V}_{{H}_{2}} = \textcolor{g r e e n}{\text{1.025 L}}$