Chemistry College

How many liters of dry H₂ are produced at STP from 8.5 g Al according to the following reaction?
2 Al (s) + 6 HCI (aq) →2 AICI3 (aq) + 3 H₂ (g)

14.1 L
7.06 L
10.6 L
21.2 L

Answers

Answer 1

The volume of H₂ gas produced at STP from 8.5 g of Al  according to the following reaction is option (3) 10.6 L

We can use the stoichiometry method to determine the volume of hydrogen gas produced from a given amount of aluminium in the following reaction:2 Al (s) + 6 HCI (aq) →2 AICI3 (aq) + 3 H₂ (g)To use the stoichiometry method, the balanced equation of the reaction is required.

From the equation, we know that the mole ratio of Al to H₂ is 2:3. This means that 2 moles of Al will produce 3 moles of H₂ gas. Therefore, we can find the number of moles of H₂ gas produced by the number of moles of Al.

Using the given mass of Al, we can find the number of moles of Al:

mass of Al = 8.5 g

Molar mass of Al = 26.98 g/mol

Number of moles of Al = mass of Al/molar mass of Al= 8.5 g/26.98 g/mol= 0.315 mol

From the mole ratio in the balanced equation, we know that 2 moles of Al will produce 3 moles of H₂.

Therefore, the number of moles of H₂ produced is:

Number of moles of H₂ = 3/2 × number of moles of Al= 3/2 × 0.315 mol= 0.473 mol

To find the volume of hydrogen gas produced, we need to use the ideal gas law equation which is PV = nRT.

Since the volume is required at STP, we can use the molar volume of gas at STP, which is 22.4 L/mol.To find the volume of H₂ gas, we can use the following formula:

Volume of H₂ gas = number of moles of H₂ × molar volume at STP= 0.473 mol × 22.4 L/mol= 10.6 L

Therefore, the volume of H₂ gas produced at STP from 8.5 g of Al is 10.6 L. The correct option is option (3) 10.6 L.

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