Question

2.0mol of (NH4)3PO4
calculate the grams of phosphate (PO4) ions

Answer 1

Solution.

$\nu((NH_4)_3PO_4) = 2.0 mol;$

$m = \nu\sdot M;$

$M(PO_4^3- ) = 31 + 16\sdot4 = 95 g/mol;$

$2.0 mol (NH_4)_3PO_4 contains 2.0mol PO_4^3-;$

$m(PO_4^3-) = 95 g/mol \sdot 2.0 mol = 190g;$

Answer: $m(PO_4^3-) = 190g.$