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Question: A \({H_3}B{O_3}\) on heating decomposes in two ways: i.\({H_3}B{O_3} \to HB{O_2} + {H_2}O\) ii.\...

A H3BO3{H_3}B{O_3} on heating decomposes in two ways:
i.H3BO3→HBO2+H2O{H_3}B{O_3} \to HB{O_2} + {H_2}O
ii.H3BO3→B2O3+H2O{H_3}B{O_3} \to {B_2}{O_3} + {H_2}O
If 9 moles of H3BO3{H_3}B{O_3} are taken, some part decomposed like (I) and remaining like (II). If total 11 moles of water are formed, the moles of B2O3{B_2}{O_3} formed is:

Explanation

Solution

To answer this question you must recall the basic fundamentals of stoichiometry. Since matter can neither be created nor destroyed, nor can one element change into the other in a chemical reaction, thus the amount of each element must be the same throughout the entire reaction. For example, the number of atoms of any element in the reactants will be always equal to the number of atoms of that element in the products formed.

Complete step by step answer:
We are given two different reactions of the decomposition of boric acid. Writing the balanced equation for both the reactions, we get,
H3BO3→HBO2+H2O{H_3}B{O_3} \to HB{O_2} + {H_2}O
2H3BO3→B2O3+3H2O2{H_3}B{O_3} \to {B_2}{O_3} + 3{H_2}O
From reaction (I), we can conclude that 1 mole of orthoboric acid gives one mole of water.
In reaction (II), we can see that two moles of orthoboric acid give three moles of water.
We are given that nine moles of orthoboric acid produce 11 moles of water on decomposition.
Taking the number of moles of orthoboric acid decomposed through reaction (I) as xx and those through reaction (II) as 2y2y.
We can write two linear equations as x+2y=9x + 2y = 9 and x+3y=11x + 3y = 11
Solving, we get, x=5x = 5 and y=2y = 2

So, the number of moles of B2O3{B_2}{O_3} formed is 2.
Note:
The heating of boric acid proceeds through three major steps. When heated mildly, at a temperature of around 370 Kelvin, orthoboric acid loses a molecule of water to form metaboric acid.
H3BO3→ΔHBO2+H2O{H_3}B{O_3}\xrightarrow[{}]{\Delta }HB{O_2} + {H_2}O
When further heated on a comparatively higher temperature, the metaboric acid forms loses water further and results into the formation of tetra boric acid. When tetra boric acid is subjected to red heat, it further undergoes loss of water to result in the formation of boron trioxide or borax.