2. If 5.4 g Al are reacted with excess HCI, how many moles of H2 will be produced? 2 Al + HCI → 2 AICI, + 3 H, (unbalanced]

0.3 moles of H2 will be produced.

Explanation:

           Balanced reaction: 2 Al + 6 HCl → 2 AlCl3 + 3 H2

Given Al mass = 5.4 g, we know Mr of Al is 27

Using the formula:

moles = mass / Mr

moles of Al = 5.4 / 27

moles of Al = 0.2 moles

Here we see that "2" molar mass before Al; so in this reaction, "1" molar mass will have [(0.2)/2] = 0.1 moles. He see "3" molar mass before H2, therefore moles of H2 is (3 * 0.1) = 0.3 moles.