A piece of pure mangnesium was completely burnt in a mixture of N_2 and O_2 and the mixture of MgO and Mg_3N_2 so obtained had a mass of 1.8g.When this mixture was heated with excess water and the...

A piece of pure mangnesium was completely burnt in a mixture of N_2 and O_2 and the mixture of MgO and Mg_3N_2 so obtained had a mass of 1.8g.When this mixture was heated with excess water and the product obtaiend ignited, only MgO was formed. The mass of this MgO was 2.0g.

Write balanced chemical equations for all the relevant reactions(ignore the reaction between MgO and H_2O) and Calculate the mole ratio MgO:Mg_3N_2 in the mixture formed by burning the piece of magnesium.(Mg = 24, O = 16 , N= 14)

1 Answer | Add Yours

llltkl's profile pic

llltkl | College Teacher | (Level 3) Valedictorian

Posted on

A piece of pure mangnesium was completely burnt in a mixture of N_2 and O_2 and the mixture of MgO and Mg_3N_2 so obtained had a mass of 1.8 g.

When this mixture was heated with excess water and the product obtaiend ignited, only MgO was formed. The mass of this MgO was 2.0g.

The balanced chemical equations for the relevant reactions are:

`2Mg+O2(g) rarr 2MgO `

`3Mg+N2(g) rarr Mg3N2 `

`Mg3N2 + 3H2O rarr 3Mg(OH)2 + 2NH3 `

`Mg(OH)2 stackrel Delta rarr MgO+H2O`

Let the x moles of mgO and y moles of Mg3N2 was formed in the initial reaction.

Thus total mass of MgO and Mg3N2 obtained (in g) is:

x(24+16)+y(3*24+2*14)

=40x+100y

By the first condition of the problem, 40x+100y=1.8 --- (i)

Again, 1 mole of Mg3N2 upon reaction with water produces three moles of Mg(OH)2 which upon heating produces three moles of MgO.

Therefore, total mass of MgO formed at the end of the second stage is:

40x+y*3*40

By the second condition of the problem, 40x+120y=2.0 --- (ii)

Solving (i) and (ii) yields, x=0.02 and y=0.01

Mole ratio MgO:Mg_3N_2 in the mixture formed by burning the piece of magnesium was 0.02:0.01, i.e. 2:1.

We’ve answered 318,926 questions. We can answer yours, too.

Ask a question