Answer: The mass of aluminium metal that must be added is 27.0 grams
Explanation:
We are given:
Volume of hydrogen gas = 33.6 L
At STP conditions:
22.4 L of volume is occupied by 1 mole of a gas
So, 33.6 L of volume will be occupied by = \(\frac{1}{22.4}\times 33.6=1.5mol\) of hydrogen gas
The chemical equation for the reaction of aluminium metal and HCl follows:
\(2Al+6HCl\rightarrow 2AlCl_3+3H_2\)
By Stoichiometry of the reaction:
3 moles of hydrogen gas is formed by 2 moles of aluminium metal
So, 1.5 moles of hydrogen gas will be formed by = \(\frac{2}{3}\times 1.5=1\) mole of aluminium metal
To calculate the number of moles, we use the equation:
\(\text{Number of moles}=\frac{\text{Given mass}}{\text{Molar mass}}\)
Moles of aluminium metal = 1 mole
Molar mass of aluminium = 27.0 g/mol
Putting values in above equation, we get:
\(1mol=\frac{\text{Mass of aluminium metal}}{27.0g/mol}\\\\\text{Mass of aluminium metal}=(1mol\times 27.0g/mol)=27.0g\)
Hence, the mass of aluminium metal that must be added is 27.0 grams
