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Standard XI

Chemistry

Stoichiometric Calculations

If 1.11 g Mg2...

Question

If 1.11 g $M g_{2} P_{2} O_{7}$ is formed at the end of the reaction, then what weight of $N a H_{2} P O_{4}$ was present originally given that no other products contained phosphorus?
$(Molar mass M g_{2} P_{2} O_{7} = 222 g m o l^{- 1}, N a H_{2} P O_{4} = 120 g m o l^{- 1})$
$N a H_{2} P O_{4} + M g^{2 +} + N H_{4}^{+} \to M g (N H_{4}) P O_{4} .6 H_{2} O \to M g_{2} P_{2} O_{7}$

2.1 g

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3.0 g

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5.0 g

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1.2 g

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Solution

The correct option is D 1.2 g
$N a H_{2} P O_{4} + M g^{2 +} + N H_{4}^{+} \to M g (N H_{4}) P O_{4} .6 H_{2} O \to M g_{2} P_{2} O_{7}$
Applying POAC on P,
Moles of P in $N a H_{2} P O_{4}$ = moles of P in $M g_{2} P_{2} O_{7}$
$1 \times number of mol of N a H_{2} P O_{4} = 2 \times number of mol of M g_{2} P_{2} O_{7}$
$\frac{W e i g h t o f N a H_{2} P O_{4}}{m o l a r m a s s} = 2 \times \frac{W e i g h t o f M g_{2} P_{2} O_{7}}{m o l a r m a s s} \frac{W e i g h t o f N a H_{2} P O_{4}}{120} = 2 \times \frac{1.11}{222}$
Weight of $N a H_{2} P O_{4}$ =1.2 g
Thus, the weight of $N a H_{2} P O_{4}$ present originally was 1.2 g

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Similar questions

Q. If 1.11 g

M g_{2} P_{2} O_{7}

is formed at the end of the reaction, then what weight of

N a H_{2} P O_{4}

was present originally given that no other products contained phosphorus?

(Molar mass M g_{2} P_{2} O_{7} = 222 {g mol}^{- 1}, N a H_{2} P O_{4} = 120 {g mol}^{- 1})

N a H_{2} P O_{4} + M g^{2 +} + N H_{4}^{+} \to M g (N H_{4}) P O_{4} .6 H_{2} O \to M g_{2} P_{2} O_{7}

Q. If 1.11 g

M g_{2} P_{2} O_{7}

is formed at the end of the reaction, then what weight of

N a H_{2} P O_{4}

was present originally given that no other products contained phosphorus?

(Molar mass M g_{2} P_{2} O_{7} = 222 g m o l^{- 1}, N a H_{2} P O_{4} = 120 g m o l^{- 1})

N a H_{2} P O_{4} + M g^{2 +} + N H_{4}^{+} \to M g (N H_{4}) P O_{4} .6 H_{2} O \to M g_{2} P_{2} O_{7}

Q. In the determination of

P

, an aqueous solution of

N a H_{2} P O_{4}

is treated with a mixture of ammonium and magnesium ions to precipitate magnesium ammonium phosphate,

M g (N H_{4}) P O_{4.} 6 H_{2} O .

This was heated and yielded

3.33

g of

M g_{2} P_{2} O_{7} .

What weight of

N a H_{2} P O_{4}

was present originally?

Q. In a gravimetric determination of P, an aqueous solution of

N a H_{2} P O_{4}

is treated with a mixture of ammonium and magnesium ions to precipitate magnesium ammonium phosphate

M g (N H_{4}) P O_{4} .6 H_{2} O

. This is heated and decomposed to magnesium pyrophosphate,

M g_{2} P_{2} O_{7}

which is weighed. A solution of

N a H_{2} P O_{4}

yielded 0.222 g of

M g_{2} P_{2} O_{7}

. What weight of

N a H_{2} P O_{4}

is present originally?

(Atomic mass:

N a = 23 u, M g = 24 u, P = 31 u, N = 14 u, O = 16 u, H = 1 u

)

Q. A certain sample of phosphate rock contains

26.26 %

P_{2} O_{5}

. A

0.5428

g sample is analysed by precipitating

M g N H_{4} P O_{4} .6 H_{2} O

and igniting the precipitate to

M g_{2} P_{2} O_{7}

. Thus,

M g_{2} P_{2} O_{7}

obtained is :

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Stoichiometric Calculations

Standard XI Chemistry

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If 1.11 g Mg2P2O7 is formed at the end of the reaction, then what weight of NaH2PO4 was present originally given that no other products contained phosphorus? (Molar mass Mg2P2O7=222 g mol−1, NaH2PO4=120g mol−1) NaH2PO4+Mg2++NH+4→Mg(NH4)PO4.6H2O→Mg2P2O7