# An iron plate weighing 5.2 g was kept in a solution of copper sulfate weighing 1.6 g. At the end of the reaction

An iron plate weighing 5.2 g was kept in a solution of copper sulfate weighing 1.6 g. At the end of the reaction, the plate was removed from the solution and dried. The weight of the plate became.

Fe + CuSO4 = FeSO4 + Cu.
1) According to the periodic table, we calculate the molar mass of copper sulfate:
64 + 32 + 16 * 4 = 160.
2) Find the amount of copper sulfate substance. To do this, divide the mass of sulfate by its molar mass: 1.6 / 160 = 0.01 mol.
3) According to the reaction equation, there is one mole of iron per mole of copper sulfate. This means that the amount of iron substance is 0.01 mol.
4) Molar mass of iron: 56.
5) Weight of iron: 0.01 * 56 = 0.56g.
6) The mass of the iron plate after the reaction is: 5.2 – 0.56 = 4.64 g – the answer.

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