Question

How many moles of oxygen gas will be formed from$6.45\,g$ of potassium chlorate?

Answer 1

Chlorate to give oxygen. Also, we should know the chemical formula for potassium chlorate. We should be quite familiar with the unitary method in order to solve these types of questions.

Formula used: $n = \dfrac{w}{M}$
Where w is the given weight in g
M is the Molecular mass in $g\,mo{l^{ - 1}}$

Complete step-by-step answer:
The balanced chemical reaction of decomposition of potassium chlorate to give oxygen is written below.
$2KCl{O_3}(s) \to 2KCl(s) + 3{O_2}(g)$
As we can see from the reaction,
Two moles of potassium chlorate will give three moles of oxygen gas.
Since we are given with the weight of potassium chlorate, hence we need to convert moles into weight to proceed further in this question.
Calculating molecular mass of potassium chlorate,
Molar mass of K (potassium) $= 39g\,mo{l^{ - 1}}$
Molar mass of Cl (chlorine) $= 35.5\,g\,mo{l^{ - 1}}$
Molar mass of O (oxygen) $= 16g\,mo{l^{ - 1}}$
Molar mass of potassium chlorate$=$ $39 + 35.5 + 3(16)$ $=$ $122.5\,g\,mo{l^{ - 1}}$
$n = \dfrac{w}{M}$
$n = \dfrac{{6.54}}{{122.5}} = 0.05\,mol$
Now using unitary method,
Two moles of potassium chlorate give three moles of oxygen gas
Hence, one mole of potassium chlorate will give $1.5$ (one and a half) moles of oxygen gas
Now, $0.05\,mol$of potassium chlorate will give $1.5 \times 0.05 = 0.075\,mol$of oxygen gas.

Hence, $0.075$mol is the required answer.
Note: Balanced chemical reaction has to be written while solving these types of questions. Stoichiometry has to be taken special care of while doing these questions in exams. Stoichiometry makes use of the coefficient ratio set up by balanced reaction equations to make connections between the reactants and products. Hence, it calculates the quantities of reactants and products.