Sulfuric acid dissolves aluminum metal according to the reaction: 2Al(s)+3H2SO4(aq)⇒Al2(SO4)3(aq)+3H2(g) Suppose you want to dissolve...

Step 1
The balanced equation for the reaction between sulfuric acid ${\displaystyle \left(H_{2}S{O}_4\right)}$ an aluminum ${\displaystyle \left(Al\right)}$ is:
${\displaystyle 2Al\left(s\right)+3H_{2}S{O}_4\left(aq\right)\Rightarrow A{l}_2{\left(S{O}_4\right)}_3\left(aq\right)+3H_2\left(g\right)}$
Based on the balanced equation, 2 mol Al reacts with 3 mol ${\displaystyle H_{2}S{O}_4}$ to form 1 mol ${\displaystyle A{l}_2{\left(S{O}_4\right)}_3}$ and 3 mol ${\displaystyle H_2}$
Step 2
We can determine the amount of ${\displaystyle H_{2}S{O}_4}$ (in g) needed to dissolve 12.7 g Al by using the mole ratio their molar masses. Molar mass of ${\displaystyle Al=26.98\frac{g}{mol}}$. Molar mass of ${\displaystyle H_{2}S{O}_4=98.08\frac{g}{mol}}$
${\displaystyle g{H}_{2}S{O}_4=12.7gAl\times \frac{molAl}{26.98g}\times \frac{3mol{H}_{2}S{O}_4}{2molAl}\times \frac{98.08g}{mol{H}_{2}S{O}_4}}$
${\displaystyle =69.3g{H}_{2}S{O}_4}$
Thus, 69.3 g ${\displaystyle H_{2}S{O}_4}$ is needed to dissolve a 12.7 g Al block
Step 3
We can determine the amount of ${\displaystyle H_2}$ (in g) produced from the reaction between 69.3 g ${\displaystyle H_{2}S{O}_4}$ and 12.7 g Al using the mole ratio and the molar mass of ${\displaystyle H_2}$ which is equal to ${\displaystyle 2.016\frac{g}{mol}}$
${\displaystyle g{H}_2=12.7gAl\times \frac{molAl}{26.98g}\times \frac{3mol{H}_2}{2molAl}\times \frac{2.016g}{mol{H}_2}}$
${\displaystyle =1.43g{H}_2}$
Therefore, 1.43 g ${\displaystyle H_2}$ is produced from the rection between 69.3 g ${\displaystyle H_{2}S{O}_4}$ and 12.7 g Al.

Step 1
${\displaystyle 2Al\left(s\right)+3H_{2}S{O}_4\left(aq\right)\Rightarrow A{l}_2\left(S{O}_4\right)3\left(aq\right)+3H_2\left(g\right)}$
Moles of ${\displaystyle Al=\frac{mass}{molar}}$ mass of Al
${\displaystyle =\frac{15.6}{26.98}=0.5782mol}$
Moles of ${\displaystyle H_{2}S{O}_4=\frac{3}{2}\times \text{moles of}Al}$
${\displaystyle =\frac{3}{2}\times 0.5782=0.8673mol}$
Mass of ${\displaystyle H_{2}S{O}_4=\text{moles}\times \text{molar mass of}{H}_{2}S{O}_4}$
${\displaystyle =0.8673\times 98.08}$
${\displaystyle =85.1g}$
Step 2
Moles of ${\displaystyle H_2=\frac{3}{2}\times \text{moles of}Al}$
${\displaystyle =\frac{3}{2}\times 0.5782=0.8673mol}$
Mass of ${\displaystyle H_2=\text{moles}\times \text{molar mass of}{H}_2}$
${\displaystyle =0.8673\times 2.016}$
${\displaystyle =1.75g}$

Step 1 $2Al(s)+3H_{2}S{O}_4(aq)\Rightarrow A{l}_2(S{O}_4{)}_3(aq)+3H_2(g)$ Moles of $Al=\frac{mass}{molarmassofAl}$ $=\frac{15.2}{26.98}=0.5634mol$ Moles os $H_{2}S{O}_4=\frac{3}{2}\times \text{moles of}Al$ $=\frac{3}{2}\times 0.5634=0.8451mol$ Mass of $H_{2}S{O}_4=\text{moles}\times \text{molar mass of}{H}_{2}S{O}_4$ $=0.8451\times 98.08=82.9g$ MSL Moles of $H_2=\frac{3}{2}\times \text{moles of}Al$ Mass os $H_2=\text{moles}\times \text{molar mass of}{H}_2$ $=0.8451\times 2.016=1.70g$