A lump of copper wire weighing 40 g was kept in a solution of mercury nitrate (2), as a result of which the weight

A lump of copper wire weighing 40 g was kept in a solution of mercury nitrate (2), as a result of which the weight of the wire increased to 45.48 g. What is the weight of the released mercury equal to?

M (Cu) = 64 g / mol.

M (Hg) = 201 g / mol.

Let’s find the changed molar mass of substances.

201 – 64 = 137 g / mol.

Let’s find the mass of the wire.

45.48 – 40 = 5.48 g

Let’s find the amount of copper substance.

n = 5.48 g: 137 g / mol = 0.04 mol.

m (Cu) = 0.04 mol × 64 g / mol = 2.56 g.

Let’s compose the reaction equation, find the quantitative ratios of the substances Cu and Hg.

Cu + Hg (NO3) 2 = Cu (NO3) 2 + Hg.

Substances are in quantitative ratios 1: 1.

n (Cu) = n (Hg) = 0.04 mol.

Find the mass Hg

M (Hg) = 201g / mol.

m (Hg) = 0.04 mol × 201 g / mol = 8.4 g.

Answer: 8.4 g.



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