Question #36e9a
1 Answer
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
magnesium and hydrogen gas. This means that the reaction will produce
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
STP conditions have been changed to
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")#