Question

The amount of mg in grams to be dissolved in dilute $H_{2} SO_{4}$ to liberate $H_{2}$ which is just sufficient to reduce 80g of ferric oxide is

Answer 1

Answer:

The balanced equations are as follows:

Mg+H

2

SO

4

→MgSO

4

+H

2

Fe

2

O

3

+3H

2

→2Fe+3H

2

O

Thus, 3 moles of magnesium corresponds to one mole of ferric oxide.

Atomic mass of magnesium, iron and oxygen are 24.3, 55.8 and 16.0 g/mol respectively.

Assuming that to reduce 160 g of ferric oxide, the amount of magnesium required will be w grams.

=>

2×55.8+3×16

160

molFe

2

O

3

×

1molFe

2

O

3

3molMg

=

24.3

w

molMg

Hence, w=

2×55.8+3×16

160

×3×24.3

=73g.

Answer 2

Answer:

The balanced equations are as follows:

Mg+H

2

SO

4

→MgSO

4

+H

2

Fe

2

O

3

+3H

2

→2Fe+3H

2

O

Thus, 3 moles of magnesium corresponds to one mole of ferric oxide.

Atomic mass of magnesium, iron and oxygen are 24.3, 55.8 and 16.0 g/mol respectively.

Assuming that to reduce 160 g of ferric oxide, the amount of magnesium required will be w grams.

=>

2×55.8+3×16

160

molFe

2

O

3

×

1molFe

2

O

3

3molMg

=

24.3

w

molMg

Hence, w=

2×55.8+3×16

160

×3×24.3

=73g.