Question

What is the experimental yield (in grams) of the solid product when the percent yield is 81.4% with 10.63 g of iron(II) nitrate reacting in solution with excess sodium phosphate?

Fe(NO₃)₂(aq) + Na₃PO₄(aq) --> Fe₃(PO₄)₂(s) + NaNO₃(aq) [unbalanced]

Answer 1

Ans. unbalanced chemical equation

Fe(NO3)2 (aq) + Na3PO4(aq) = Fe3(PO4)2 (s) + NaNO3

Now, the balanced equation is:-

3Fe(NO3)2 (aq) + 2Na3PO4(aq) = Fe3(PO4)2 (s) + 6NaNO3

molar mass of iron(II) nitrate = 180 g/mol

molar mass of iron(II) phosphate = 357.48 g/mol

3 * 180 g of iron(II) nitrate gives 357.48 g of iron(II) phosphate

10.63 g of iron (II) nitrate gives (357.48 * 10.63) /540 = theoretical yield = 7.03 g iron(II) phosphate

% yield = (experimental yield / theoretical yield ) * 100 = 81.4

experimental yield = (81.4 * 7.03) / 100 = 5.72 g