The reaction equation has the form:
Fe2O3 + 3H2SO4 = Fe2 (SO4) 3 + 3H2O
From the reaction equation it follows that the formation of 1 mole of iron sulfate (lll) requires 3 moles of sulfuric acid, we get:
n [Fe2 (SO4) 3] = 3n (H2SO4) or n [Fe2 (SO4) 3]: 3n (H2SO4);
M (H2SO4) = 98 g / mol; M [Fe2 (SO4) 3] = 400 g / mol.
We calculate the mass and amount of sulfuric acid using data from the conditions of the problem, we obtain:
m (H2SO4) = m (solution) * w% = (245 g * 60%) / 100% = 147 g;
n (H2SO4) = m (H2SO4) / M (H2SO4) = 147 g / 98 g / mol = 1.5 mol.
From here we find the mass of iron sulfate (lll):
1 mol [Fe2 (SO4) 3]: 3 mol (H2SO4) = chol [Fe2 (SO4) 3]: 1.5 mol (H2SO4);
chol [Fe2 (SO4) 3] = [1 mol [Fe2 (SO4) 3] * 1.5 mol (H2SO4)] / 3 mol (H2SO4) = 0.5 mol [Fe2 (SO4) 3].
m [Fe2 (SO4) 3] = n [Fe2 (SO4) 3] * M [Fe2 (SO4) 3] = 0.5 mol * 400 g / mol = 200 g.
Answer: m [Fe2 (SO4) 3] = 200g
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