Aluminum metal reacts with hcl to produce aluminum chloride and hydrogen gas. how many grams of aluminum metal must be added to an excess of hcl to produce 33.6 l of hydrogen gas, if the gas is at stp?
(a) 18.0 g
(b) 35.0 g
(c) 27.0 g
(d) 4.50 g
(e) 9.00 g

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 = of hydrogen gas

The chemical equation for the reaction of aluminium metal and HCl follows:

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 = mole of aluminium metal

To calculate the number of moles, we use the equation:

Moles of aluminium metal = 1 mole

Molar mass of aluminium = 27.0 g/mol

Putting values in above equation, we get:

Hence, the mass of aluminium metal that must be added is 27.0 grams

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