Question

Find the value of a, b, c, and d in accordance with the law of conservation of mass for the reaction given below.
$aM{g}_3{N}_2\left(s\right)+b{H}_{2}O\left(l\right)\to cMg\left(OH{)}_2\right(aq)+dN{H}_3(aq)$

• A. a, b, c and d are 1, 3, 4 and 2 respectively.
• B. a, b, c and d are 2, 6, 3 and 2 respectively.
• C. a, b, c and d are 1, 6, 3 and 2 respectively.
• D. a, b, c and d are 1, 6, 2 and 3 respectively.

Answer 1

The correct option is C

a, b, c and d are 1, 6, 3 and 2 respectively.

$aM{g}_3{N}_2\left(s\right)+b{H}_{2}O\left(l\right)\to cMg\left(OH{)}_2\right(aq)+dN{H}_3(aq)$

For magnesium number of atoms on the reactant and

product side will be:- 3a = c

For nitrogen, it will be:- 2a = d

For hydrogen, it will be:- 2b =2c + 3d

For oxygen it will be:- b = 2c

Now take a=1 and we will this value in by 3a = c to get c = 3, and in 2a =d

to get d = 2 and after putting value of c in b = 2c we will get b = 6

The value of a, b, c, and d are 1, 6, 3 and 2 respectively in accordance with the

law of conservation of mass. Law of conservation of mass states that mass can neither be created

nor be destroyed and hence the number of atoms on the left and right side of the equation must be

equal for the mass to be same. So the resultant reaction after balancing would be:-

$M{g}_3{N}_2\left(s\right)+6H_{2}O\left(l\right)\to 2Mg\left(OH{)}_2\right(aq)+2N{H}_3(aq)$ and so the

number of atoms on left and right side are equal.