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

reaction: 2Al + 6HCl ----> 2AlCl3 + 3H2

Mass of Al = 27g/mole
Mass of H2 = 2g/mole = 22.4dm3 (STP)

1l = 1dm3

according to the reaction:
2*27g of Al ---------------------- 3*22,4dm3 H2
x g of Al -------------------------- 33,6dm3 H2
x = 27.0g of Al

answer: C

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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 = [tex]\frac{1}{22.4}\times 33.6=1.5mol[/tex] of hydrogen gas

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

[tex]2Al+6HCl\rightarrow 2AlCl_3+3H_2[/tex]

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 = [tex]\frac{2}{3}\times 1.5=1[/tex] mole of aluminium metal

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

[tex]\text{Number of moles}=\frac{\text{Given mass}}{\text{Molar mass}}[/tex]

Moles of aluminium metal = 1 mole

Molar mass of aluminium = 27.0 g/mol

Putting values in above equation, we get:

[tex]1mol=\frac{\text{Mass of aluminium metal}}{27.0g/mol}\\\\\text{Mass of aluminium metal}=(1mol\times 27.0g/mol)=27.0g[/tex]

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