A mixture weighing 20 g, consisting of copper and copper oxide (2 valence) was treated

| January 31, 2021 | education

A mixture weighing 20 g, consisting of copper and copper oxide (2 valence) was treated with an excess of concentrated nitric acid. At the same time, gas was collected with a volume of 1.568 dm3. Calculate the mass fractions of substances in the original mixture.

Given: m (Cu + CuO) = 20 g

+ HNO3 (conc.)

V (gas) = ​​1.568 cubic decimetres or 1,568 l

w (Cu) =? and W (CuO) =?

Decision.

Copper reacts with concentrated nitric acid to release gas. Let’s write the reaction equation:

Cu + 4HNO3 = Cu (NO3) 2 + 2NO2 + 2H2O

Molar mass (copper) = 64 g / mol, Molar volume (under normal conditions) is 22.4 l / mol.

Let’s make the ratio: 64 g (Cu) emits 2 * 22.4 L (NO2)

x g (Cu) emits 1.568 L (NO2), hence

x = 64 * 1.58 / 44.8 = 2.24 Gg (Cu)

Let’s calculate the mass fraction of copper in the mixture: w (Cu) = 2.24 * 100% / 20 = 11.2%.

Let’s calculate the mass fraction of copper (II) oxide:

w (CuO) = 100% – 11.2% = 88.8%

Answer: the mass fraction of copper in the mixture is 11.2%, the mass fraction of copper (II) oxide is 88.8%



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