WinterPlakke778
Answered

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



Answer :

pyromorphite
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

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

Other Questions