Question

In: Chemistry

What is the percent yield of the solid product when 17.62 g of iron(III) nitrate reacts...

What is the percent yield of the solid product when 17.62 g of iron(III) nitrate reacts in solution with excess sodium phosphate and 5.863 g of the precipitate is experimentally obtained? Fe(NO3)3(aq) + Na3PO4(aq) --> FePO4(s) + NaNO3(aq) [unbalanced]

Expert Solution

Given reaction, Fe(NO₃)₃(aq) + Na₃PO₄(aq) --> FePO₄(s) + NaNO₃(aq)

Balanced reaction is Fe(NO₃)₃(aq) + Na₃PO₄(aq) --> FePO₄(s) + 3 NaNO₃(aq)

Precipitate = FePO₄(s)

molar mass of Fe(NO₃)₃ = 241.8 g/mol

molar mass of FePO4 = 150.8 g/mol

Fe(NO₃)₃(aq) + Na₃PO₄(aq) --> FePO₄(s) + 3 NaNO₃(aq)

1 mol 1 mol

241.8 g 150.8 g

17.62 g ?

From the balanced equation,it is clear that

241.8 g of Fe(NO₃)₃ produce150.8 g of FePO4.

So, 17.62 g of Fe(NO₃)₃ produce _? g of FePO4.

? =[ 17.62 g of Fe(NO₃)₃/241.8 g of Fe(NO₃)₃] x 150.8 g of FePO4

= 10.99 g of FePO4

So, 17.62 g of Fe(NO₃)₃ produce 10.99 g of FePO4.

This is called theoretical yield.

Given that actual yield of FePO4 = 5.863 g

Hence,

percent yield = (actual yield of FePO4/ theoretical yield of FePO4) x 100

= (5.863 g /10.99 g) x 100

= 53.35 %

Therefore, percent yield of the solid product FePO4 = 53.35 %

queen_honey_blossom answered 1 hour ago

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