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m_a_m_a [10]
3 years ago
15

If 4.0 mol aluminum and 7.0 mol hydrogen bromide react according to the following equation, how many moles of hydrogen are forme

d and what is the limiting reactant?
Chemistry
1 answer:
Vinil7 [7]3 years ago
3 0

Answer:

Moles of hydrogen formed = 3.5 moles

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

<u>Moles of hydrogen formed = 3.5 moles</u>

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