Question

calculate the mass of KClO3 required to produced 6.72 litre of O2 at S.T.P[K=39 , Cl=35.5, O=16]

Answer 1

Answer:

$\boxed{24.5 g \:of \:KClO3\: is \:required \:to \:produce\: 6.72\:l\: of \:O2 \:Gas\: at \:STP.}$

Explanation:

$\huge \text{Given}$

Atomic Mass of :-

$\bigstar$ Potassium(K) = 39 u

$\bigstar$ Chlorine(Cl) = 35.5 u

$\bigstar$ Oxygen (O) = 16 u

$\huge \text{To Find}$

$The\ mass\ of\ KClO3\ required\ to \ produced\ 6.72 \ litre \ of \ O2 \ at \ S.T.P$.

$\huge \text{Answer}$

$\sf 2KCIO_3 \xrightarrow{MnO_2} 2KCI+3O_2$

We know that 1 mole of gas occupies 22.4 l volume at STP.

So , Number of Moles Required To Produce 6.72 l of Oxygen = 6.72/ 22.4 = 0.3 moles

From the Chemical Equation Mentioned Above:-

$\bigstar$ We can Observe That 2 moles of KClO₃​ will produce 3 moles of O₂ .

$\bigstar$ 0.3 moles of oxygen gas is produced by  2/3 x 0.3 = 0.2 moles of KClO₃

$\bigstar$ Mass of 1 mole of ​ = 39+ 35.5+3 × 16 = 122.5 g

∴ The mass of  KClO₃ required = 0.2 × 122.5 = 24.5 g.