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 partial pressure 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.
