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Question

What weights of P4O6 and P4O10 will be produced by the combustion of 31 g of P4 in 32 g of oxygen leaving no P4 and no O2?


A

55 g and 71 g

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B

17.5 g and 19.5 g

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C

2.75 g and 35.5 g

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D

2.75 g and 29.5 g

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Solution

The correct option is C

2.75 g and 35.5 g


Explanation for the correct option:

Step 1: Write the balanced chemical reaction and molar mass of P4 and O2.

P4 (s)+3O2(g)P4O6(s)

Molar mass of P4=124 g/mol

Molar mass of O2=32 g/mol

Step 2: Calculate the moles of P4 and O2.

Moles of P4=31 g124 g/mol=0.25 moles

Moles of O2=32 g32 g/mol=1 mole

Step 3: Determine the limiting reagent in the reaction.

As 1 mole of P4 reacts with 3 moles of O2

Therefore,

0.25 moles of P4 will react with 0.25×3 moles=0.75 moles of O2

Thus, P4 is the limiting reagent which is completely consumed in the reaction and Oxygen is left.

Step 4: Determine the moles of P4O6

Since 1 mole of P4 produces 1 mole of P4O6, therefore, 0.25 moles of P4 produces 0.25 moles of P4O6.

Step 5: Write the reaction of combustion of P4O6 and find the limiting reagent of the reaction.

P4O6 (s)+2O2(g)P4O10(s)

As 1 mole of P4O6 reacts with 2 moles of O2

Therefore,

0.25 moles of P4O6 will react with 0.25×2 moles=0.50 moles of O2

The amount of Oxygen required for this reaction is not available as 0.75 moles O2 is already consumed and only 0.25 moles O2 is left, Oxygen is the limiting reagent.

Step 6: Calculate the moles of P4O10 produced.

Since 2 moles of O2 produces 1 mole of P4O10, therefore, 0.25 moles of O2 produces 0.252=0.125 moles of P4O10.

Number of moles of P4O10 left =0.25-0.125=0.125 moles

Step 7: Calculate the weight of P4O6 and P4O10

Mass of P4O6=0.125 mol×220 g/mol=27.5 g

Mass of P4O6=0.125 mol×220 g/mol=27.5 g

Hence, the correct option is (C).

Explanation for the incorrect options:

Since, the correct weights of P4O6 and P4O10 are 2.75 g and 35.5 g. Therefore, options (A), (B) and (D) stand incorrect.

Therefore, the correction option is (C).


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