Question

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]

Answer 1

Balanced chemical equation is given by

`\\ \tt\hookrightarrow 2Al+HCl\longrightarrow 2AlCl_3+3H_2`

Moles of Al:-

`\\ \tt\hookrightarrow 5.4/27=0.2mol`

Now

• 2mol of Al produces 3moles H_2.

• 1mol of Al produces 1.5mol H_2
• 0.2mol Al produces 0.3 mol H_2

Answer 2

0.3 moles of H2 will be produced.

Step-by-step 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.