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Limiting Reagent or Reactant

Which of the ...

Question

Which of the following statements is/are wrong for the following reactions?

$P_{4} + 5 O_{2} \to P_{4} O_{10}$
$P_{4} + 3 O_{2} \to P_{4} O_{6}$

$1.24$ g of $P_{4}$ reacts with $8.0$ g of $O_{2}$ .

A

P_{4}

is the limiting quantity

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B

O_{2}

is the limiting quantity

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C

Mass of

P_{4} O_{10}

obtained is

2.2

g

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D

Mass of

P_{4} O_{6}

obtained is

2.84

g

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Solution

The correct options are
B $O_{2}$ is the limiting quantity
C Mass of $P_{4} O_{10}$ obtained is $2.2$ g
D Mass of $P_{4} O_{6}$ obtained is $2.84$ g
$n_{P_{4}} = \frac{1.24}{31 \times 4} = 0.01 m o l, n_{O_{2}} = \frac{8}{32} = 0.25 m o l$

In reaction (i),
Moles of $O_{2}$ required $= (0.01 m o l P_{4}) (\frac{5 m o l O_{2}}{m o l P_{4}})$

$= 0.01 \times 5 = 0.05 m o l$

Since $O_{2}$ is in excess amount.

Therefore, $P_{4}$ is the limiting quantity.

Molecular weight $P_{4} O_{10} = 284 g$

0.01 mol of $P_{4}$ produces $=$ 0.01 mol of $P_{4} O_{10}$

So $P_{4} O_{10}$ in terms of mass $= 0.01 \times 284 = 2.84 g$

Molecular weight of $P_{4} O_{6} = 220$

Weight of $P_{4} O_{6}$ produced $= (0.01 m o l P_{4}) (\frac{1 m o l P_{4} O_{6}}{m o l P_{4} O_{6}}) \times M_{P_{4} O_{6}}$
$= 0.01 \times 220 = 2.2 g$

Hence options B,C & D are correct.

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Similar questions

Q. Two substance,

P_{4}

O_{2}

are allowed to react completely to form mixture of

P_{4} O_{6}

P_{4} O_{10}

leaving none of the reactants. The composition of final mixture if

P_{4}

O_{2}

P_{4} + 3 O_{2} \to P_{4} O_{6}

P_{4} + 5 O_{2} \to P_{4} O_{10}

.

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P_{4} O_{10}

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P_{4}

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Q. A reaction container holds 5.77 g of

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and the following reaction occurs:

P_{4} + O_{2} \to P_{4} O_{6} .

The amount of excess reagent left is?

Q. What weights of

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440

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