Question

Question #36e9a

Answer 1

`"1.025 L"`

Explanation:

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

`"Mg"_text((s]) + 2"HCl"_text((aq]) -> "MgCl"_text(2(aq]) + "H"_text(2(g]) uarr`

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 `"22.4 L"`.

STP conditions have been changed to `"100 kPa"` and `0^@"C"`, but that doesn't change the fact that at `"1 atm"` and `0^@"C"`, one mole of any ideal gas occupies `"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) = color(green)("1.025 L")`