Question

s) Suppose we now collect hydrogen gas, H2(g), over water at 21◦C in a vessel with total pressure of 743 Torr. If the hydrogen gas is produced by the reaction of aluminum with hydrochloric acid: 2Al(s) + 6HCl(aq) → 2AlCl3(aq) + 3H2(g) what volume of hydr

Answer 1

This is an incomplete question, here is a complete question.

Suppose we now collect hydrogen gas, H₂(g), over water at 21°C in a vessel with total pressure of 743 Torr. If the hydrogen gas is produced by the reaction of aluminum with hydrochloric acid:

`2Al(s)+6HCl(aq)\rightarrow 2AlCl_3(aq)+3H_2(g)`

what volume of hydrogen gas will be collected if 1.35 g Al(s) reacts with excess HCl(aq)? Express your answer in liters.

Answer : The volume of hydrogen gas that will be collected is 1.85 L

Explanation :

First we have to calculate the number of moles of aluminium.

Given mass of aluminium = 1.35 g

Molar mass of aluminium = 27 g/mol

`\text{Number of moles}=\frac{\text{Given mass}}{\text{Molar mass}}`

`\text{Moles of aluminium}=(1.35g)/(27g/mol)=0.05mol`

The given chemical reaction is:

`2Al(s)+6HCl(aq)\rightarrow 2AlCl_3(aq)+3H_2(g)`

As, hydrochloric acid is present in excess. So, it is considered as an excess reagent.

Thus, aluminium is a limiting reagent because it limits the formation of products.

By Stoichiometry of the reaction:

2 moles of aluminium produces 3 moles of hydrogen gas

So, 0.005 moles of aluminium will produce =
`(3)/(2)* 0.05=0.0750mol` of hydrogen gas

Now we have to calculate the mass of helium gas by using ideal gas equation.

PV = nRT

where,

P = Pressure of hydrogen gas = 743 Torr

V = Volume of the helium gas = ?

n = number of moles of hydrogen gas = 0.075 mol

R = Gas constant =
`62.364\text{ L Torr }mol^(-1)K^(-1)`

T = Temperature of hydrogen gas =
`21^oC=[21+273]K=294K`

Now put all the given values in above equation, we get:

`743Torr* V=0.075mol* 62.364\text{ L Torr }mol^(-1)K^(-1)* 294K\\\\V=1.85L`

Hence, the volume of hydrogen gas that will be collected is 1.85 L