Question

Write balanced equation for:
(i) $B{{F}_{3}}+LiH\to $
(ii) ${{B}_{2}}{{H}_{6}}+{{H}_{2}}O\to $
(iii) $NaH+{{B}_{2}}{{H}_{6}}\to $
(iv) ${{H}_{3}}B{{O}_{3}}\xrightarrow{\Delta }$
(v) $Al+NaOH\to $
(vi) ${{B}_{2}}{{H}_{6}}+N{{H}_{3}}\to $

Answer 1

Hint: The required answer to this question lies in the fact that the number of moles of each atom on the reactant side should be equal to that on the product side. Based on this find the products of the given reactions and then balance on both sides.

Complete step by step answer:
In the previous classes of chemistry which deal with the very basic chemistry, we have come across the concept of balancing the given chemical equation and how to interpret it.
We shall now refresh the concept so that we will approach the required answer.
- A chemical equation is the interpretation of the compound that reacts together to form a new or modified compound.
- Balancing this chemical equation means that the number of moles of each atom on the reactant side should be equal to that of the product side.
- Now let us see each reaction in detail and balance accordingly.
(i) In this, the reactants together produce diborane and lithium fluoride. The balancing of this equation will give the answer,
 $2B{{F}_{3}}+6LiH\to {{B}_{2}}{{H}_{6}}+6LiF$
(ii) In this reaction, diborane reacts with water to yield orthoboric acid and the balanced equation will be,
${{B}_{2}}{{H}_{6}}+6{{H}_{2}}O\to 2{{H}_{3}}B{{O}_{3}}$
(iii) Here, sodium hydride on reaction with diborane gives sodium borohydride which is the reducing agent. The balanced reaction will be,
$2NaH+{{B}_{2}}{{H}_{6}}\to 2NaB{{H}_{4}}$
(iv) Orthoboric acid on heating will decompose to give metaboric acid with the release of the water molecules. The balanced reaction is,
${{H}_{3}}B{{O}_{3}}\xrightarrow{\Delta }HB{{O}_{2}}+{{H}_{2}}O$
This metaboric acid on further heating can oriduce boron trioxide that is,
$4HB{{O}_{2}}\xrightarrow[-{{H}_{2}}O]{\Delta }{{H}_{2}}{{B}_{4}}{{O}_{7}}\xrightarrow{\Delta }2{{B}_{2}}{{O}_{3}}(borontrioxide)+{{H}_{2}}O$
(v) Aluminium on reaction with a base that is sodium hydroxide reacts to give aluminum hydroxide with the release of sodium.
$Al+3NaOH\to Al{{(OH)}_{3}}+3Na$
(vi) Diborane when reacted with ammonia and in the presence of heat will give out hydrogen gas. The reaction is,
$3{{B}_{2}}{{H}_{6}}+6N{{H}_{3}}\to 2{{B}_{3}}{{N}_{3}}{{H}_{6}}+12{{H}_{2}}$

Note: Note that the balancing of a chemical equation need not be necessarily in integer form but can also be in the fraction form. The above reaction can also be written in the fraction balanced form.