Question

Write a balanced equation for
(i) Silicon dioxide is treated with hydrogen fluoride
(ii) Boric acid is added to water
(iii) Diborane reacts with \[N{H_3}\] followed by heating.

Answer 1

Hint: A balanced equation is an equation which has an equal number of atoms on both sides of the equation. The balancing of the equation is governed by the law of conservation of mass.

Complete step by step answer:
A balanced equation has the same number of atoms on both sides of reactants and products. The total mass of reactants is equal to the masses of products by combining the masses of elements and molecules on either side. This is known as the law of conservation of mass.
Let us consider the given reactions one by one.

(i) Silicon dioxide is treated with hydrogen fluoride.
Silicon dioxide has the molecular formula \[Si{O_2}\] and hydrogen fluoride has the molecular formula \[HF\]. Then the product has to identify by the chemical reaction of the starting materials. The products are \[Si{F_4}\] and \[{H_2}O\]. Thus the unbalanced equation is
$Si{O_2} + HF \to Si{F_4} + {H_2}O$
Monitoring the equation clearly indicates that the equivalent of \[HF\] is to be multiplied by\[4\]and the equivalent of \[{H_2}O\] is to be multiplied by\[2\]. Thus the balanced equation is
$Si{O_2} + 4HF \to Si{F_4} + 2{H_2}O$

(ii) Boric acid is added to water
Boric acid has the molecular formula \[{H_3}B{O_3}\] and water has the molecular formula \[{H_2}O\]. Then the product of the reaction is \[{\left[ {B{{\left( {OH} \right)}_4}} \right]^ - }\] and \[{H_3}{O^ + }\]. The boric acid accepts electrons from \[O{H^ - }\] and releases protons. Thus the unbalanced equation is
${H_3}B{O_3} + {H_2}O \to {[B{(OH)_4}]^ - } + {H_3}{O^ + }$
Monitoring the equation clearly indicates that the equivalent of \[{H_2}O\] is to be multiplied by \[2\]. Also the charge has to be neutral on both sides. Thus the balanced equation is
${H_3}B{O_3} + 2{H_2}O \to {[B{(OH)_4}]^ - } + {H_3}{O^ + }$

(iii) Diborane reacts with \[N{H_3}\] followed by heating.
Diborane has the molecular formula \[{B_2}{H_6}\] and \[N{H_3}\] is ammonia. The product of the reaction of diborane with \[N{H_3}\] in presence of heating gives borazine and hydrogen. The unbalanced equation of the reaction is
${B_2}{H_6} + N{H_3} \to {B_3}{N_3}{H_6} + {H_2}$
Monitoring the equation clearly indicates that the equivalent of diborane has to be multiplied by \[3\] and the equivalent of \[N{H_3}\] have to be multiplied by \[6\]. In the product side the equivalent of borazine has to be multiplied by \[2\] and the equivalent of \[{H_2}\] has to be multiplied by \[12\]. Thus the balanced chemical equation is
$3{B_2}{H_6} + 6N{H_3} \to 2{B_3}{N_3}{H_6} + 12{H_2}$

Note:
A balanced chemical equation is essential to predict the exact stoichiometry of the reactants and products. The reaction also leads to evaluating the exact yield of a reaction and percentage yield of a reaction.