Calculation of Volume of H2 Evolved from Reaction of 0.9g Al with Dil. HCl at STP

Given:

Mass of Al (m) = 0.9g

Molar Mass of Al (M) = 27g/mol

2Al(s) + 6HCl(aq) → 2AlCl3(aq) + 3H2(g)

The balanced chemical equation shows that 3 moles of H2 are produced for every 2 moles of Al reacted.

Calculations:

1. Moles of Al reacted = m/M = 0.9/27 = 0.0333 mol
2. Moles of H2 produced = 3/2 x 0.0333 = 0.05 mol
3. Volume of H2 at STP = (22.4 L/mol) x 0.05 mol = 1.12 L

Explanation:

The given problem is a stoichiometry problem that requires the use of the balanced chemical equation to determine the moles of H2 produced from the reaction of 0.9g Al with dilute HCl. The balanced chemical equation shows that 3 moles of H2 are produced for every 2 moles of Al reacted. Therefore, the first step is to calculate the moles of Al reacted using its mass and molar mass.

Once the moles of Al reacted are known, the moles of H2 produced can be determined using the mole ratio from the balanced chemical equation. Finally, the volume of H2 at STP can be calculated using the molar volume of a gas (22.4 L/mol) and the number of moles of H2 produced.

Conclusion:

The volume of H2 evolved from the reaction of 0.9g Al with dilute HCl at STP is 1.12 L.

1.12 liters maybe
tell me is it right or not