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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