180 g of 10% iron sulfate (3) was mixed with 100 g of 4% sodium hydroxide solution. The precipitate that formed

180 g of 10% iron sulfate (3) was mixed with 100 g of 4% sodium hydroxide solution. The precipitate that formed was heated and filtered. What is the mass of the substance remaining on the filter?

Given:
m solution (Fe2 (SO4) 3) = 180 g
ω (Fe2 (SO4) 3) = 10%
m solution (NaOH) = 100 g
ω (NaOH) = 4%

Find:
m (in the filter) -?

Solution:
1) Fe2 (SO4) 3 + 6NaOH => 2Fe (OH) 3 ↓ + 3Na2SO4;
2Fe (OH) 3 = (toC) => Fe2O3 + 3H2O;
2) m (Fe2 (SO4) 3) = ω (Fe2 (SO4) 3) * m solution (Fe2 (SO4) 3) / 100% = 10% * 180/100% = 18 g;
3) n (Fe2 (SO4) 3) = m (Fe2 (SO4) 3) / M (Fe2 (SO4) 3) = 18/400 = 0.045 mol;
4) m (NaOH) = ω (NaOH) * m solution (NaOH) / 100% = 4% * 100/100% = 4 g;
5) n (NaOH) = m (NaOH) / M (NaOH) = 4/40 = 0.1 mol;
6) n (Fe (OH) 3) = n (NaOH) * 2/6 = 0.1 * 2/6 = 0.033 mol;
7) n (Fe2O3) = n (Fe (OH) 3) / 2 = 0.033 / 2 = 0.0165 mol;
8) m (Fe2O3) = n (Fe2O3) * M (Fe2O3) = 0.0165 * 160 = 2.64 g.

Answer: The mass of Fe2O3 is 2.64 g.



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