Question

How much potassium bromide KBr is produced if 20.0 g of potassium bromate, $KBr{{O}_{3}}$ decomposes?

Answer 1

Hint: This problem is completely based on the balanced chemical equation and the stoichiometry it describes. A balanced chemical equation describes the stoichiometric coefficients which can further be used to solve many real-life problems.

Complete step-by-step answer: Let us solve this problem in detail;
First, we need to write the balanced chemical equation which shows the conversion of potassium bromate to potassium bromide;
$2KBr{{O}_{3}}_{\left( s \right)}\xrightarrow{\Delta }2KB{{r}_{\left( s \right)}}+3{{O}_{2\left( g \right)}}$
This is the decomposition reaction where potassium bromate will undergo thermal decomposition to form potassium bromide and oxygen gas.
Here, we can see that 2 moles of potassium bromate will form 2 moles of potassium bromide. Thus, the mole ratio will be 1 : 1.
Now, using basics of stoichiometry,
167 g/mol potassium bromate will form 119 g/mol potassium bromide. Then 20 g of potassium bromate will;
Potassium bromide formed $=20.0gKBr{{O}_{3}}\times \dfrac{119.0gKBr}{167.0gKBr{{O}_{3}}}=14.3g$

Note: Do note that the problem does not contain much dimensional characteristics but we need to take them into account even if they are not very diverse.
Sometimes we don’t use stoichiometric coefficients but instead we use mole ratio; both are the same. The ratio of the stoichiometric coefficients is known as mole ratio.