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Standard XII

Chemistry

Percentage Composition

If we conside...

Question

If we consider $100 %$ yield, then number of moles of $H_{3} P O_{4}$ would be obtained is :

1.25

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10.0

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5.0

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2.5

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Solution

The correct option is B $10.0$
$310 g P_{4} = \frac{310}{124} m o l P_{4} = 2.5 m o l P_{4} ⟹ 2.5 \times 4 = 10 m o l H_{3} P O_{4}$ .

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Similar questions

Q. Phosphoric acid (

H_{3} P O_{4}

) is widely used to make fertilizers and can be prepared in a two-step process.

S t e p I : P_{4} + 5 O_{2} ⟶ P_{4} O_{10}

S t e p I I : P_{4} O_{10} + 6 H_{2} O ⟶ 4 H_{3} P O_{4}

We allow

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g of phosphorus to react with excess oxygen, which forms tetraphosphorus deoxide (

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25 %

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(Atomic weight in amu of

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Assuming

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Q. For the reaction,

F e (N O_{3})_{3} + N a_{2} C O_{3} \to F e_{2} (C O_{3})_{3} + N a N O_{3}

Initially if 2.5 moles of

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Initially if

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Q. The number of moles of

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Q. For the reaction,

2 X + 3 Y + 4 Z \to 5 W

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If we consider 100% yield, then number of moles of H3PO4 would be obtained is :

The correct option is B 10.0310gP4=310124molP4=2.5molP4⟹2.5×4=10molH3PO4.

If we consider $100 %$ yield, then number of moles of $H_{3} P O_{4}$ would be obtained is :

The correct option is B $10.0$
$310 g P_{4} = \frac{310}{124} m o l P_{4} = 2.5 m o l P_{4} ⟹ 2.5 \times 4 = 10 m o l H_{3} P O_{4}$ .