Question

One way in which the useful metal copper is produced is by dissolving the mineral azurite, which contains copper(II) carbonate, in concentrated sulfuric acid. The sulfuric acid reacts with the copper(II) carbonate to produce a blue solution of copper(II) sulfate. Scrap iron is then added to this solution, and pure copper metal precipitates out because of the following chemical reaction: Fe(s) + CuSO4(aq) + Cu(s) + FeSO4(aq) Suppose an industrial quality-control chemist analyzes a sample from a copper processing plant in the following way. He adds powdered iron to a 350. mL copper(II) sulfate sample from the plant until no more copper will precipitate. He then washes, dries, and weighs the precipitate, and finds that it has a mass of 78. mg. Calculate the original concentration of copper(II) sulfate in the sample. Be sure your answer has the correct number of significant digits. ya x 6 ?

Answer 1

Given

Volume of copper sulfate solution = 350.0 ml

Mass of solid copper = 78.0 mg

We have relation, 1 g = 1000 mg .

$\therefore$ 1 g / 1000 mg = 1 = 1000 mg / 1 g

$\therefore$ Mass of solid Cu = 78.0 mg ( 1 g / 1000 mg ) = 0.0780 g

Consider a reaction, Fe (s) + CuSO ₄ (aq) $\rightarrow$ Cu (s) + FeSO ₄ (aq)

From reaction, 1 mol Cu $\equiv$ 1 mol CuSO ₄

$\therefore$ 63.54 g Cu $\equiv$ 159.60 g CuSO ₄

$\therefore$ 0.0780 g Cu $\equiv$ 159.60 $\times$ 0.0780 / 63.54 g CuSO ₄

0.0780 g Cu $\equiv$ 0.1959 g CuSO ₄

i e 0.0780 g solid Cu is produced from 0.1959 g CuSO ₄.

Volume of CuSO ₄ solution is 350.0 ml

350.0 ml CuSO ₄ solution contain 0.1959 g CuSO ₄

$\therefore$ 1000 ml CuSO ₄ solution contain 1000 ml $\times$ ( 0.1959 g / 350.0 ml) = 0.560 g

Original concentration of copper (II) sulfate solution = 0.560 g / L