Question

For the production of 36 grams of water, 4 grams of hydrogen is required according to the following reaction 2H2+O2->2H2O.
What is the mass of oxygen required for the production of 0.72 grams of water?

Answer 1

solution:

Moles of water = 0.72/18 = 0.04 moles

Now, according to stichometery,

1 moles of O2 gives 2 mole of water.

Then, similarly 0.04 moles of water required 0.02 moles of oxygen.

So, required mass of oxygen= 16 × 0.02 = 0.32 g

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Amrit⭐

Answer 2

☆☆ AS-SALAMU-ALYKUM ☆☆

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BALANCED CHEMICAL EQUATION IS :-
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2H2 + O2 -------> 2H2O


MEANS 1 MOLE OXYGEN PRODUCES TWO MOLES OF WATER .

NOW MASS OS WATER IS GIVEN WE HAVE TO CALCULATE MOLES OF WATER

MOLAR MASS OF WATER IS = H2O = 1×2+16=18g/mol

MOLES OF WATER = MASS÷MOLAR MASS

MASS OF WATER IS GIVEN THAT IS EQUAL TO 0.72g

MOLES = 0.72 ÷ 18

MOLES OF WATER = 0.04mol

FROM BALANCED CHEMICAL EQUATION IT IS CLEAR THAT 1 MOLE OXYGEN PRODUCES 2 MOLES WATER

THEN 0.02 MOLES OXYGEN WILL PRODUCE 0.04 MOLES OF WATER
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SO FROM THIS WE GOT MOLES OF OXYGEN THAT IS 2 MOLES

LAST STEP IS TO FIND MASS OF OXYGEN :-
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MOLES OF OXYGEN = MASS ÷ MOLAR MASS

MOLAR MASS OF O2= 16×2 =32g/mol

MASS OF OXYGEN = MOLES× MOLAR MAS

MASS OF OXYGEN = 0.02 × 32

MOLES OF OXYGEN = 0.64 g

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