# Question 8c029

Oct 21, 2017

Here's what I got.

#### Explanation:

The balanced chemical equation that describes this reaction

$3 {\text{H"_ 2"SO"_ (4(aq)) + 2"Al"_ ((s)) -> "Al"_ 2("SO"_ 4)_ (3(aq)) + 3"H}}_{2 \left(g\right)} \uparrow$

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.

$14.236 \textcolor{red}{\cancel{\textcolor{b l a c k}{\text{moles H"_2"SO"_4))) * "98.079 g"/(1color(red)(cancel(color(black)("mole H"_2"SO"_4)))) = color(darkgreen)(ul(color(black)(1.40 * 10^3color(white)(.)"g}}}}$

$9.491 \textcolor{red}{\cancel{\textcolor{b l a c k}{\text{moles Al"))) * "26.982 g"/(1color(red)(cancel(color(black)("mole Al")))) = color(darkgreen)(ul(color(black)("256 g}}}}$

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