NaCl Fe(NO3)3 0.456 g NaOH Fe(OH)3 0.107 g
A mixture contains only NaCl and Fe(NO3)3.
A 0.456 g sample of the mixture is dissolved in water,
and an excess of NaOH is added,
producing a precipitate of Fe(OH)3.
The precipitate is filtered, dried, and weighed.
Its mass is 0.107 g. Calculate the following.
a. the mass of iron in the sample
b. the mass of Fe(NO3)3 in the sample
c. the mass percent of Fe(NO3)3 in the sample
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Fe(OH)3의 몰질량 = 106.87 g/mol 이므로,
0.107 g / (106.87 g/mol) = 0.0010012 mol
( 참고 https://ywpop.tistory.com/7738 )
= 0.00100 mol Fe(OH)3
Fe(OH)3 : Fe = 1 : 1 몰수비 이고,
Fe의 몰질량 = 55.85 g/mol 이므로,
0.00100 mol × (55.85 g/mol) = 0.05585 g
= 0.0559 g Fe
Fe(NO3)3(aq) + 3NaOH(aq) → Fe(OH)3(s) + 3NaNO3(aq)
Fe(NO3)3 + 3NaOH → Fe(OH)3 + 3NaNO3
( 참고 https://ywpop.tistory.com/11299 )
Fe(NO3)3 : Fe(OH)3 = 1 : 1 계수비(= 몰수비) 이고,
Fe(NO3)3의 몰질량 = 241.86 g/mol 이므로,
0.00100 mol × (241.86 g/mol) = 0.242 g Fe(NO3)3
(0.242 g / 0.456 g) × 100 = 53.1%
---> mass percent of Fe(NO3)3
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