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1 Answers

Balancing the equation we get,

3 NH4NO3 + Na3PO4 = (NH4)3PO4 + 3 NaNO3

Molecular weight of NH4NO3 is 80.043 g/mol

So, converting it to moles we get,

(30.0 g NH4NO3) / (80.043 NH4NO3 g/mol)

= 0.37480 mol NH4NO3

Molecular weight of Na3PO4 is 163.941 g/mol

So, converting it to moles we get,

(50.0 g Na3PO4) / (163.941 Na3PO4 g/mol)

= 0.30499 mol Na3PO4

0.37480 mol of NH4NO3 would react completely with 0.37480 * (1/3) mol of Na3PO4

= 0.12493 mol of Na3PO4

But the amount of Na3PO4 is more than that of NH4NO3. So, Na3PO4 is in excess.

Mass of products: (NH4)3PO4 and NaNO3:

(0.37480 mol of NH4NO3) * 1 mol (NH4)3PO4 / 3 mol NH4NO3) * (149.086 (NH4)3PO4 g/mol) = 18.6 g (NH4)3PO4

(0.37480 mole of NH4NO3) * (3 mol NaNO3 / 3 mol NH4NO3) * (84.9947 NaNO3 g/mol) = 31.9 g NaNO3

Mass of excess left over:

{(0.30499 mol Na3PO4 initially) – (0.12493 mol Na3PO4)} * (163.9407 Na3PO4 g/mol) = 29.5 g Na3PO4 will be left over.