Question #36e9a

1 Answer
Nov 3, 2015

Answer:

#"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")#