An aq soln of $N a H_{2} P O_{4}$ $NaH_2PO_4$ is treated with a mixture of ammonium & magnesium ions to precipitate $M g (N H_{4}) P O_{4} . 6 H_{2} O$ $Mg(NH_4)PO_4. 6H_2O$ & is heated to $M g_{2} P_{2} O_{7}$ $Mg_2P_2O_7$ which is weighed. A soln of $N a H_{2} P O_{4}$ $NaH_2PO_4$ gives 1.054 g of $M g_{2} P_{2} O_{7}$ $Mg_2P_2O_7$ . What's original weight of $N a H_{2} P O_{4}$ $NaH_2PO_4$ ?

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An aq soln of $N a H_{2} P O_{4}$ $NaH_2PO_4$ is treated with a mixture of ammonium & magnesium ions to precipitate $M g (N H_{4}) P O_{4} . 6 H_{2} O$ $Mg(NH_4)PO_4. 6H_2O$ & is heated to $M g_{2} P_{2} O_{7}$ $Mg_2P_2O_7$ which is weighed. A soln of $N a H_{2} P O_{4}$ $NaH_2PO_4$ gives 1.054 g of $M g_{2} P_{2} O_{7}$ $Mg_2P_2O_7$ . What's original weight of $N a H_{2} P O_{4}$ $NaH_2PO_4$ ?

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Answer 1 · 2015-07-17T17:37:52

The original mass of ${NaH}_{2} {PO}_{4}$ $"NaH"_2"PO"_4$ was 1.136 g.

Explanation:

It is tempting to write out the balanced equations for the reactions, but you can avoid doing this.

You know that phosphorus atoms must be conserved, so the overall reaction must be

${2NaH}_{2} {PO}_{4} + \dots \to {Mg}_{2} P_{2} O_{7} + \dots$ $"2NaH"_2"PO"_4 + … → "Mg"_2"P"_2"O"_7 + …$

Now we can do the usual stoichiometry calculations and convert

${grams of Mg}_{2} P_{2} O_{7} \to {moles of Mg}_{2} P_{2} O_{7} \to {moles of NaH}_{2} {PO}_{4} \to {grams of NaH}_{2} {PO}_{4}$ $"grams of Mg"_2"P"_2"O"_7 → "moles of Mg"_2"P"_2"O"_7 → "moles of NaH"_2"PO"_4 → "grams of NaH"_2"PO"_4$

$1.054 cancel("g Mg₂P₂O₇") × (1 cancel("mol Mg₂P₂O₇"))/(222.55 cancel("g Mg₂P₂O₇")) × (2 cancel("mol NaH₂PO₄"))/(1 cancel("mol Mg₂P₂O₇")) × ("119.98 g NaH"_2"PO"_4)/(1 cancel("mol NaH₂PO₄")) = "1.136 g NaH"_2"PO"_4$

Addendum

The balanced equations are

$"2×(NaH"_2"PO"_4 + "NH"_4^+ + "Mg"^(2+) + "6H"_2"O" → cancel("Mg(NH₄)PO₄·6H₂O") + "Na"^+ + "2H"^+)$
$1×(cancel("2Mg(NH₄)PO₄·6H₂O") → "Mg"_2"P"_2"O"_7 + "2NH"_3 + "13H"_2"O")$
$bar("2NaH"_2"PO"_4 + "2NH"_4^+ + "2Mg"^(2+) → "Mg"_2"P"_2"O"_7 + "2NH"_3 + "2Na"^+ + "4H"^+ + "H"_2"O")$

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