Question #52665

Q- 3 g C^2H^4 gas requires oxygen gas ( O^2) for combustion which is obtained from pure Potassium Chlorate ( KClO^3). Mass of Potassium Chlorate is ____ ?

Expert's answer

C_{2}H_{4} + 3O_{2} = 2CO_{2}+2H_{2}O

From the equation, 1 mole of C_{2}H_{4 }requires 3 moles of O_{2 }for combustion.

Consequently 3 g of C_{2}H_{4}requires X grams of O_{2}

X = 3·3/1 = 9 (grams of O_{2})

2KClO_{3 }= 2KCl + 3O_{2}

From the equation, 3 moles of O_{2 }are produced from 2 moles of Potassium Chlorate,

So 9 grams of O_{2} areproduced from Y grams of Potassium Chlorate

Y = 9·2/3 = 6 (grams of KClO_{3})

**Answer: 6 grams**

From the equation, 1 mole of C

Consequently 3 g of C

X = 3·3/1 = 9 (grams of O

2KClO

From the equation, 3 moles of O

So 9 grams of O

Y = 9·2/3 = 6 (grams of KClO

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