Question

If 4.0 mol aluminum and 7.0 mol hydrogen bromide react according to the following equation, how many moles of hydrogen are formed and what is the limiting reactant?

Answer 1

3.5 mol H2, HBr (limiting reactant)

Step-by-step explanation:

4.0 mol Al × 3 mol H2/ 2 mol Al = 6.0 mol H2

7.0 mol HB ×3 mol H2/ 6mol HBr = 3.5 mol H2

Since 7.0mol of HBr will produce less H2 than 4.0mol of Al, HBr will be the limiting reactant, and the reaction will produce 3.5mol of H2.

Answer 2

Moles of hydrogen formed = 3.5 moles

Step-by-step explanation:

Given that:-

Moles of aluminium= 4.0 mol

Moles of hydrogen bromide = 7.0 mol

According to the reaction:-

`2Al_((s))+6HBr_((aq))\rightarrow 2AlBr_3_((aq))+3H_2_((g))`

2 moles of aluminum react with 6 moles of hydrogen bromide

1 mole of aluminum react with 6/2 moles of hydrogen bromide

4 moles of aluminum react with (6/2)*4 moles of hydrogen bromide

Moles of hydrogen bromide = 12 moles

Available moles of hydrogen bromide = 7.0 moles

Limiting reagent is the one which is present in small amount. Thus, hydrogen bromide is limiting reagent. (7.0 < 12)

The formation of the product is governed by the limiting reagent. So,

6 moles of hydrogen bromide on reaction forms 3 moles of hydrogen

1 moles of hydrogen bromide on reaction forms 3/6 moles of hydrogen

7 moles of hydrogen bromide on reaction forms (3/6)*7 moles of hydrogen

Moles of hydrogen formed = 3.5 moles