Question

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?

Answer 1

The first step is to calculate the number of moles of `Al` present, `n=0.27` `mol`, then use the mole ratio in the equation to determine that this releases `0.40` `mol` 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` `Al(s)` + `3` `H_2SO_4(aq) to` `Al_2(SO_4)_3(aq)` + `3` `H_2(g)`

First calculate the number of moles of `Al` in `7.25` `g`, knowing the molar mass of `Al` is `27` `gmol^-1`:

`n=m/M=7.25/27=0.27` `mol`

From the equation we can see that each `2` `mol` of `Al` yields `3` `mol` of `H_2`, so multiply this by `3/2` to find the number of moles of `H_2` released.

`n=0.40` `mol`

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` `atm`, so this number of moles will produce `0.4xx22.4=9.0` `L` of gas at STP.

We are not asked about STP, though, but `25^o` `C` (`273` `K`) and `1.65` `atm`.

Using the combined gas law:

`(P_1V_1)/T_1=(P_2V_2)/T_2`

Rearranging to make `V_2` the subject:

`V_2=(P_1V_1)T_2/T_1P_2=(1*9*298)/(273*1.65)=5.95` `L`