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Stoichiometric Calculations

In a gravimet...

Question

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

0.24

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0.12

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0.18

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0.36

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Solution

The correct option is A $0.24$ g
P atom remains conserved in the reaction according to POAC.

Molar mass of $M g_{2} P_{2} O_{7}$ $= [24 \times 2 + 2 \times 31 + 7 \times 16] = 222 g / m o l$

Mass of $M g_{2} P_{2} O_{7} = 0.222$ g

Moles of $M g_{2} P_{2} O_{7} = \frac{0.222}{222} = 0.001 m o l$

Moles of phosphorous $= 2 \times 0.001 = 0.002 m o l = Moles of N a H_{2} P O_{4}$

Mass of $N a H_{2} P O_{4}$ $= 0.002 \times [23 + 2 + 31 + 4 \times 16]$
$= 0.002 \times 120 = 0.24 g$

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

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