Question #8c029

The balanced chemical equation that describes this reaction

#3"H"_ 2"SO"_ (4(aq)) + 2"Al"_ ((s)) -> "Al"_ 2("SO"_ 4)_ (3(aq)) + 3"H"_ (2(g)) uarr#

tells you that sulfuric acid and aluminium react in a #3:2# mole ratio. Moreover, for every #3# moles of sulfuric acid and #2# moles of aluminium that take part in the reaction, you get #3# moles of hydrogen gas.

In your case, the reaction produced

#28.7 color(red)(cancel(color(black)("g"))) * "1 mole H"_2/(2.016color(red)(cancel(color(black)("g")))) = "14.236 moles H"_2#

This implies that the reaction consumed

#14.236 color(red)(cancel(color(black)("moles H"_2))) * ("3 moles H"_2"SO"_4)/(3color(red)(cancel(color(black)("moles H"_2)))) = "14.236 moles H"_2"SO"_4#

and

#14.236 color(red)(cancel(color(black)("moles H"_2))) * "2 moles Al"/(3color(red)(cancel(color(black)("moles H"_2)))) = "9.491 moles Al"#

Finally, to convert the number of moles of each reactant to moles, use the molar masses of the two compounds.

#14.236 color(red)(cancel(color(black)("moles H"_2"SO"_4))) * "98.079 g"/(1color(red)(cancel(color(black)("mole H"_2"SO"_4)))) = color(darkgreen)(ul(color(black)(1.40 * 10^3color(white)(.)"g")))#

#9.491 color(red)(cancel(color(black)("moles Al"))) * "26.982 g"/(1color(red)(cancel(color(black)("mole Al")))) = color(darkgreen)(ul(color(black)("256 g")))#

The answers are rounded to three sig figs, the number of sig figs you have for the mass of hydrogen gas.