You visited us 1 times! Enjoying our articles? Unlock Full Access!

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

Right on! Give the BNAT exam to get a 100% scholarship for BYJUS courses

0.12

No worries! We‘ve got your back. Try BYJU‘S free classes today!

0.18

No worries! We‘ve got your back. Try BYJU‘S free classes today!

0.36

No worries! We‘ve got your back. Try BYJU‘S free classes today!

Open in App

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

Suggest Corrections

Similar questions

N a H_{2} P O_{4}

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

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

N a H_{2} P O_{4}

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

N a H_{2} P O_{4}

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

P

N a H_{2} P O_{4}

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

3.33

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

N a H_{2} P O_{4}

P_{1}

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

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

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

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

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

N a H_{2} P O_{4}

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

N a H_{2} P O_{4}

(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}

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

N a H_{2} P O_{4}

(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}

Join BYJU'S Learning Program