Question

. Aluminum nitrate can be made by the following reaction
FeCl3(aq) + 3HNO3 Fe(NO3)3 + 3HCl
A 18.5 g of HNO3 (Mwt= 63) was mixed with excess FeCl3 to obtain 21.25g of Fe(NO3)3
(Mwt=242) . Calculate the percentage yield

Answer 1

$n(HNO_3) = \frac{18.5}{63}=0.293 \;mol \\ n(Fe(NO_3)_3) = n(HNO_3)=0.293 \;mol \\ m(Fe(NO_3)_3) = 0.293 \times 242 = 71.06 \;g$

Proportion:

71.06 – 100 %

21.25 – x

$x = \frac{21.25 \times 100}{71.06}=29.9$ %

Answer: 30 %