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Standard XII

Chemistry

Stoichiometric Calculations

In the determ...

Question

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?

3.6

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3.9

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Solution

The correct option is A $3.6$ g
Weight of $P$ in $3.33$ g of $M g_{2} P_{2} O_{7}$ $= 62 \times \frac{3.33}{222} = 0.93$ g.
As this $P$ is present in reactant, so mass of reactant $= 120 \times \frac{0.93}{31} = 3.6$ g.

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Similar questions

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. In a gravimetric determination of

P_{1}

an aqueous solution of dihydrogen phosphate ion

H_{2} P {O_{4}}^{-}

is treated with the 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 weighted. A solution of

H_{2} P {O_{4}}^{-}

yeilded 1.054 g of

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

What weight of

N a H_{2} P O_{4}

what 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. 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. 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}

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In the determination of P, an aqueous solution of NaH2PO4 is treated with a mixture of ammonium and magnesium ions to precipitate magnesium ammonium phosphate, Mg(NH4)PO4.6H2O. This was heated and yielded 3.33 g of Mg2P2O7. What weight of NaH2PO4 was present originally?

The correct option is A 3.6 gWeight of P in 3.33 g of Mg2P2O7 =62×3.33222=0.93 g.As this P is present in reactant, so mass of reactant =120×0.9331=3.6 g.

The correct option is A $3.6$ g
Weight of $P$ in $3.33$ g of $M g_{2} P_{2} O_{7}$ $= 62 \times \frac{3.33}{222} = 0.93$ g.
As this $P$ is present in reactant, so mass of reactant $= 120 \times \frac{0.93}{31} = 3.6$ g.