# Indicate the amount of substances in water that is formed during the thermal decomposition of ferrum

Indicate the amount of substances in water that is formed during the thermal decomposition of ferrum (III) hydroxide weighing 42.8 g.

Given:
m (Fe (OH) 3) = 42.8 g

Find:
n (H2O) -?

Solution:
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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