Question

An aq soln of `NaH_2PO_4` is treated with a mixture of ammonium & magnesium ions to precipitate `Mg(NH_4)PO_4. 6H_2O` & is heated to `Mg_2P_2O_7` which is weighed. A soln of `NaH_2PO_4` gives 1.054 g of `Mg_2P_2O_7`. What's original weight of `NaH_2PO_4`?

Answer 1

The original mass of `"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 + … → "Mg"_2"P"_2"O"_7 + …`

Now we can do the usual stoichiometry calculations and convert

`"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")`