Question

compute the mass in grams of kico3 necessary to produce 67.2 litres of oxygen at STP according to the reaction=
KCIO3 - - - -> 2KCL + 3O2
(given= molecular weight of KCIO3=122.5)

explain fuLl process
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nd u will get 100 score​

Answer 1

Answer:

245 grams

Explanation:

According to avogadro's law, 1 mole of every substance occupies 22.4 Liters at STP and contains avogadro's number   6.023 × 10²³ of particles.

2 KClO₃ ⇒ 2 KCl +  3 O₂

According to stoichiometry, 3 moles of  O₂  are produced from 2 moles of KClO₃

Thus  3 * 22.4 = 67.2 L of  O₂  are produced from 2 moles of KClO₃

Mass of  KClO₃ =  2 × 122.5 = 245 g

Thus 245 grams of  KClO₃  necessary to produce 67.2 liters of oxygen at   S. T. P.

Answer 2

HOLA MATE

ANSWER:-

At STP , 1 mol gas = 22.4L . Therefore 67.2L = 67.2/22.4 = 3 mol O2.

3 mol O2 = 32*3 = 96g O2

1mol CacO3 will contain :

Molar mass O = 16g/mol : 3O = 48g

There is 48g O contained in 1 mol CaCO3

Therefore 96g O2 will be contained in 2 mol CaCO3

Molar mass CaCO3 = 100g/mol

Mass of 2 mol CacO3 = 200g

Answer: 200g CacO3 will contain 67.2L O2 at STP .