# When 7.25 g of Al reacts with excess H_2SO_4 in the reaction 2Al (s) + 3 H_2SO_4 (aq) -> Al_2(SO_4)_3(aq) + 3H_2(g) how many liters of H_2 is formed at 25° C and a pressure of 1.650 atm?

Mar 10, 2016

The first step is to calculate the number of moles of $A l$ present, $n = 0.27$ $m o l$, then use the mole ratio in the equation to determine that this releases $0.40$ $m o l$ of ${H}_{2}$, with a volume of $9.0$ $L$ at STP. Next convert to the given temperature and pressure, and $V = 5.95$ $L$.

#### Explanation:

$2$ $A l \left(s\right)$ + $3$ ${H}_{2} S {O}_{4} \left(a q\right) \to$ $A {l}_{2} {\left(S {O}_{4}\right)}_{3} \left(a q\right)$ + $3$ ${H}_{2} \left(g\right)$

First calculate the number of moles of $A l$ in $7.25$ $g$, knowing the molar mass of $A l$ is $27$ $g m o {l}^{-} 1$:

$n = \frac{m}{M} = \frac{7.25}{27} = 0.27$ $m o l$

From the equation we can see that each $2$ $m o l$ of $A l$ yields $3$ $m o l$ of ${H}_{2}$, so multiply this by $\frac{3}{2}$ to find the number of moles of ${H}_{2}$ released.

$n = 0.40$ $m o l$

We know that a mole of an ideal gas occupies $22.4$ $L$ at standard temperature and pressure (STP), which is ${0}^{o}$ $C$ or $273$ $K$ and $1$ $a t m$, so this number of moles will produce $0.4 \times 22.4 = 9.0$ $L$ of gas at STP.

We are not asked about STP, though, but ${25}^{o}$ $C$ ($273$ $K$) and $1.65$ $a t m$.

Using the combined gas law:

$\frac{{P}_{1} {V}_{1}}{T} _ 1 = \frac{{P}_{2} {V}_{2}}{T} _ 2$

Rearranging to make ${V}_{2}$ the subject:

${V}_{2} = \left({P}_{1} {V}_{1}\right) {T}_{2} / {T}_{1} {P}_{2} = \frac{1 \cdot 9 \cdot 298}{273 \cdot 1.65} = 5.95$ $L$