osnomu3

2022-01-14

Sulfuric acid dissolves aluminum metal according to the reaction:
$2Al\left(s\right)+3{H}_{2}S{O}_{4}\left(aq\right)⇒A{l}_{2}{\left(S{O}_{4}\right)}_{3}\left(aq\right)+3{H}_{2}\left(g\right)$
Suppose you want to dissolve an aluminum block with a mass of 15.2 g.
What minimum mass of ${H}_{2}S{O}_{4}$ (in g) do you need? What mass of ${H}_{2}$ gas (in g) does the complete reaction of the aluminum block produce?

Virginia Palmer

Expert

Step 1
The balanced equation for the reaction between sulfuric acid $\left({H}_{2}S{O}_{4}\right)$ an aluminum $\left(Al\right)$ is:
$2Al\left(s\right)+3{H}_{2}S{O}_{4}\left(aq\right)⇒A{l}_{2}{\left(S{O}_{4}\right)}_{3}\left(aq\right)+3{H}_{2}\left(g\right)$
Based on the balanced equation, 2 mol Al reacts with 3 mol ${H}_{2}S{O}_{4}$ to form 1 mol $A{l}_{2}{\left(S{O}_{4}\right)}_{3}$ and 3 mol ${H}_{2}$
Step 2
We can determine the amount of ${H}_{2}S{O}_{4}$ (in g) needed to dissolve 12.7 g Al by using the mole ratio their molar masses. Molar mass of $Al=26.98\frac{g}{mol}$. Molar mass of ${H}_{2}S{O}_{4}=98.08\frac{g}{mol}$

Thus, 69.3 g ${H}_{2}S{O}_{4}$ is needed to dissolve a 12.7 g Al block
Step 3
We can determine the amount of ${H}_{2}$ (in g) produced from the reaction between 69.3 g ${H}_{2}S{O}_{4}$ and 12.7 g Al using the mole ratio and the molar mass of ${H}_{2}$ which is equal to $2.016\frac{g}{mol}$

Therefore, 1.43 g ${H}_{2}$ is produced from the rection between 69.3 g ${H}_{2}S{O}_{4}$ and 12.7 g Al.

godsrvnt0706

Expert

Step 1
$2Al\left(s\right)+3{H}_{2}S{O}_{4}\left(aq\right)⇒A{l}_{2}\left(S{O}_{4}\right)3\left(aq\right)+3{H}_{2}\left(g\right)$
Moles of $Al=\frac{mass}{molar}$ mass of Al
$=\frac{15.6}{26.98}=0.5782mol$
Moles of ${H}_{2}S{O}_{4}=\frac{3}{2}×\text{moles of}Al$
$=\frac{3}{2}×0.5782=0.8673mol$
Mass of ${H}_{2}S{O}_{4}=\text{moles}×\text{molar mass of}{H}_{2}S{O}_{4}$
$=0.8673×98.08$
$=85.1g$
Step 2
Moles of ${H}_{2}=\frac{3}{2}×\text{moles of}Al$
$=\frac{3}{2}×0.5782=0.8673mol$
Mass of ${H}_{2}=\text{moles}×\text{molar mass of}{H}_{2}$
$=0.8673×2.016$
$=1.75g$

alenahelenash

Expert

Step 1 $2Al\left(s\right)+3{H}_{2}S{O}_{4}\left(aq\right)⇒A{l}_{2}\left(S{O}_{4}{\right)}_{3}\left(aq\right)+3{H}_{2}\left(g\right)$ Moles of $=\frac{15.2}{26.98}=0.5634mol$ Moles os ${H}_{2}S{O}_{4}=\frac{3}{2}×\text{moles of}Al$ $=\frac{3}{2}×0.5634=0.8451mol$ Mass of ${H}_{2}S{O}_{4}=\text{moles}×\text{molar mass of}{H}_{2}S{O}_{4}$ $=0.8451×98.08=82.9g$ MSL Moles of ${H}_{2}=\frac{3}{2}×\text{moles of}Al$ Mass os ${H}_{2}=\text{moles}×\text{molar mass of}{H}_{2}$ $=0.8451×2.016=1.70g$