Indicate the amount of substances in water that is formed during the thermal decomposition of ferrum (III) hydroxide weighing 42.8 g.
m (Fe (OH) 3) = 42.8 g
n (H2O) -?
1) Write the reaction equation:
2Fe (OH) 3 => Fe2O3 + 3H2O;
2) Calculate the molar mass of Fe (OH) 3:
M (Fe (OH) 3) = Mr (Fe (OH) 3) = Ar (Fe) * N (Fe) + Ar (O) * N (O) + Ar (H) * N (H) = 56 * 1 + 16 * 3 + 1 * 3 = 107 g / mol;
3) Calculate the amount of substance Fe (OH) 3:
n (Fe (OH) 3) = m (Fe (OH) 3) / M (Fe (OH) 3) = 42.8 / 107 = 0.4 mol;
4) Determine the amount of substance H2O:
n (H2O) = n (Fe (OH) 3) * 3/2 = 0.4 * 3/2 = 0.6 mol.
Answer: The amount of H2O substance is 0.6 mol.
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