# Answer to Question #52665 in Inorganic Chemistry for Tajnur

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

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

_{2}H_{4}+ 3O_{2}= 2CO_{2}+2H_{2}OFrom 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 ChlorateY = 9·2/3 = 6 (grams of KClO

_{3})**Answer: 6 grams**Need a fast expert's response?

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