We're asked to find the number of grams of $"Al"$ that reacted, given some $"H"_2 (g)$ product characteristics.
Let's first write the chemical equation for this reaction:
$2"Al"(s) + 6"HCl"(aq) rarr2 "AlCl"_3(aq) + 3"H"_2 (g)$
The total pressure of the gaseous system is given as $751$ $"mm Hg"$, and the of water vapor is $26.8$ $"mm Hg"$ at $27^"o""C"$. The pressure of hydrogen gas is thus
$P_"total" = P_ ("H"_2"O") + P_ ("H"_2)$
$P_ ("H"_2) = 751$ $"mm Hg"$ $- 26.8$ $"mm Hg"$ $= color(red)(724$ $color(red)("mm Hg"$
This pressure in atmospheres is
$724cancel("mm Hg")((1color(white)(l)"atm")/(760cancel("mm Hg"))) = color(red)(0.953$ $color(red)("atm"$
We'll now use the ideal gas equation to find the number of moles of $"H"_2$ formed:
($T = 27^"o""C" + 273 = 300$ $"K"$)
$n = (PV)/(RT) = ((color(red)(0.953)cancel(color(red)("atm")))(0.0465cancel("L")))/((0.082057(cancel("L")•cancel("atm"))/("mol"•cancel("K")))(300cancel("K"))) = color(green)(0.00180$ $color(green)("mol H"_2$
(volume converted to liters here)
Using the coefficients of the chemical equation, we'll now find the relative number of moles of $"Al"$ that react:
$color(green)(0.00180)cancel(color(green)("mol H"_2))((2color(white)(l)"mol Al")/(3cancel("mol H"_2))) = color(purple)(0.00120$ $color(purple)("mol Al"$
Lastly, we'll use the molar mass of aluminum ($26.98$ $"g/mol"$) to find the number of grams that reacted:
$color(purple)(0.00120)cancel(color(purple)("mol Al"))((26.98color(white)(l)"g Al")/(1cancel("mol Al"))) = color(blue)(0.0324$ $color(blue)("g Al"$
Thus, $color(blue)(0.0324$ $sfcolor(blue)("grams of aluminum"$ reacted.
