Suppose we now collect hydrogen gas, h2(g), over water at 21◦c in a vessel with total pressure of 743 torr. if the hydrogen gas is produced by the reaction of aluminum with hydrochloric acid:
2al(s) + 6hcl(aq) → 2alcl3(aq) + 3h2(g)
what volume of hydrogen gas will be collected if 1.35 g al(s) reacts with excess hcl(aq)? express
your answer in liters.

Explanation:

1. Gather all the information in one place

M_r:   26.98

            2Al + 6HCl ⟶ 1AlCl₃ + 3H₂

m/g:     1.35

T = 21 °C

2. Moles of Al

3. Moles of H₂

The molar ratio is 3 mol H₂: 2 mol Al

4. Volume of H₂

We can use the Ideal Gas Law to calculate the volume of hydrogen.

The gas it collected over water, so the gas collected is a mixture of hydrogen and water vapour.

T = 21 + 273.15 = 294.15 K

answer: mole of the gas

answer: mole of the gasexplanation:

the answer is a! (2.9 x 10^-18 j) hope i could someone. : )