2 Fe(OH)3 + 3 H2SO4 "\\rightarrow" Fe2(SO4)3 + 6 H2O
As you can see from the balanced equation, 2 molecules of Fe(OH)3 react with 3 molecules of H2SO4. According to this stoichiometry, the number of the moles that are consumed relate as:
"\\frac{n(Fe(OH)_3)}{2} = \\frac{n(H_2SO_4)}{3}"
Let's evaluate the left and the right side of this equation separately, using the quantity of the substances given:
"n(Fe(OH)_3)\/2 = \\frac{4.2}{2} = 2.1"
"n(H_2SO_4)\/3 = \\frac{6.5}{3} = 2.17"
2.17>2.1, so Fe(OH)3 is the limiting reactant. In other words, there is an excess of H2SO4.
Answer: For the conditions given, the limiting reactant is Fe(OH)3.
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