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Chapter 20: Problem 4

Write a balanced equation to represent the roasting of copper(I) sulfide to form "blister copper"

Short Answer

Expert verified
Answer: Cu2S (s) + O2 (g) ⟶ 2 Cu (s) + SO2 (g)

Step by step solution

01

Write the unbalanced chemical equation

First, we need to write the unbalanced equation representing the process: Cu2S (s) + O2 (g) ⟶ Cu (s) + SO2 (g)
02

Count the elements on both sides

In the unbalanced equation we see that there are: Left-hand side (reactants): 2 Cu, 1 S, 2 O Right-hand side (products): 1 Cu, 1 S, 2 O
03

Balance the elements

To make sure that the numbers of atoms of each element are equal on both sides, we will balance the equation by adjusting the stoichiometric coefficients (the numbers in front of the compounds). First, we need to balance the copper (Cu) atoms. Since there are 2 Cu atoms in Cu2S and only 1 Cu atom in Cu, we need to multiply the Cu on the right-hand side by 2: Cu2S (s) + O2 (g) ⟶ 2 Cu (s) + SO2 (g) Now, the equation reads: Left-hand side: 2 Cu, 1 S, 2 O Right-hand side: 2 Cu, 1 S, 2 O The balanced equation is: Cu2S (s) + O2 (g) ⟶ 2 Cu (s) + SO2 (g)

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Most popular questions from this chapter

Taking \(K_{\text {ip }} \mathrm{PbCl}_{2}=1.7 \times 10^{-5}\) and assuming \(\left[\mathrm{Cl}^{-}\right]=0.20 \mathrm{M}\), calculate the concentration of \(\mathrm{Pb}^{2+}\) at equilibrium.

Iron(II) can be oxidized to iron(III) by permanganate ion in acidic solution. The permanganate ion is reduced to manganese(II) ion. (a) Write the oxidation half-reaction, the reduction half-reaction, and the overall redox equation. (b) Calculate \(E^{\circ}\) for the reaction. (c) Calculate the percentage of Fe in an ore if a \(0.3500-\mathrm{g}\) sample is dissolved and the \(\mathrm{Fe}^{2+}\) formed requires for titration \(55.63 \mathrm{~mL}\) of a \(0.0200 \mathrm{M}\) solution of \(\mathrm{KMnO}_{4}\)

Write a balanced redox equation for the reaction of mercury with aqua regia, assuming the products include \(\mathrm{HgCl}_{4^{2-}}\) and \(\mathrm{NO}_{2}(g)\).

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