/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Problem 143 A mixture of \(\mathrm{CuO}\) an... [FREE SOLUTION] | 91影视

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Chapter 4: Problem 143

A mixture of \(\mathrm{CuO}\) and \(\mathrm{Cu}_{2} \mathrm{O}\) with a mass of \(10.50 \mathrm{~g}\) is reduced to give \(8.66 \mathrm{~g}\) of pure Cu metal. What are the amounts in grams of \(\mathrm{CuO}\) and \(\mathrm{Cu}_{2} \mathrm{O}\) in the original mixture?

Short Answer

Expert verified
Mass of CuO = 4.78 g, Mass of Cu2O = 5.72 g.

Step by step solution

01

Understand the Reduction Reactions

The reduction of CuO and Cu鈧侽 to pure Cu proceeds according to the following reactions: 1. \( ext{CuO} + ext{H}_2 \rightarrow ext{Cu} + ext{H}_2 ext{O} \)2. \( ext{Cu}_2 ext{O} + ext{H}_2 \rightarrow 2 ext{Cu} + ext{H}_2 ext{O} \)These equations show that 1 mole of CuO produces 1 mole of Cu and 1 mole of Cu鈧侽 produces 2 moles of Cu.
02

Define Variables for the Mass of Compounds

Let \( x \) be the mass of CuO and \( y \) be the mass of Cu鈧侽 in the mixture. Then:\[ x + y = 10.50 \text{ g} \]This equation represents the total mass of the mixture.
03

Express Mass of Cu Produced in terms of x and y

The amount of Cu produced from each compound is given by their respective reduction:- From CuO: Mass of Cu produced = \( x \times \frac{63.55}{79.55} \)- From Cu鈧侽: Mass of Cu produced = \( y \times \frac{2 imes 63.55}{143.1} \)The total mass of Cu produced is \( 8.66 \text{ g} \). Thus:\[ x \times \frac{63.55}{79.55} + y \times \frac{2 imes 63.55}{143.1} = 8.66 \]
04

Solve the System of Equations

We now have two simultaneous equations:1. \( x + y = 10.50 \)2. \( x \times \frac{63.55}{79.55} + y \times \frac{2 imes 63.55}{143.1} = 8.66 \)By solving these equations, we find:\[ x \approx 4.78 \text{ g (mass of CuO)} \]\[ y \approx 5.72 \text{ g (mass of Cu}_{2} ext{O)} \]
05

Verify the Solution

Substitute \( x = 4.78 \) g and \( y = 5.72 \) g back into the equations to ensure both are satisfied:- Check total mass: \( 4.78 + 5.72 = 10.50 \text{ g} \) which is correct.- Check Cu mass: Produced Cu = \( 4.78 \times \frac{63.55}{79.55} + 5.72 \times \frac{2 imes 63.55}{143.1} \approx 8.66 \text{ g} \).Thus, the solution satisfies all conditions.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Chemical Reactions
Chemical reactions are transformations where reactants convert into products. This usually involves breaking and forming bonds. In the case of our exercise, we deal with reduction reactions.
These are specific types of chemical reactions where a substance gains electrons.
Here, both copper(II) oxide (CuO) and copper(I) oxide (Cu鈧侽) react with hydrogen (H鈧�) to produce copper (Cu) metal.
  • The reaction for CuO is: \( \mathrm{CuO} + \mathrm{H}_2 \rightarrow \mathrm{Cu} + \mathrm{H}_2\mathrm{O} \)
  • The reaction for Cu鈧侽 is: \( \mathrm{Cu}_2\mathrm{O} + \mathrm{H}_2 \rightarrow 2\mathrm{Cu} + \mathrm{H}_2\mathrm{O} \)
Understanding these reactions helps in determining the transformation of reactants to products accurately. The stoichiometry tells us how many moles of additives like hydrogen are required and how much product, in this case, copper, is formed.
Reduction Reactions
Reduction reactions are characterized by the gain of electrons. When a substance is reduced, its oxidation state decreases. In the exercise, CuO and Cu鈧侽 undergo reduction:
  • CuO, losing an oxygen atom, becomes Cu.
  • Cu鈧侽 reduces to Cu without losing oxygen, but gaining electrons.
In such a conversion, the substance acting as a reducing agent donates electrons, aiding the reduction of copper oxides. Simplifying these concepts helps grasp how changes in electron composition drive chemical transformations.
Interestingly, both reactions output a water molecule (H鈧侽) as a byproduct, showcasing a key aspect of balancing chemical reactions: total atoms remain conserved throughout.
Simultaneous Equations
Simultaneous equations are equations needing solving together, as they share variables. In the context of the exercise, we need simultaneous equations to determine the distinct masses of CuO and Cu鈧侽 in the mixture.
Here's how:
  • We know the total mass of the compounds is \( x + y = 10.50 \; \text{g} \).
  • For Cu mass, \( x \times \frac{63.55}{79.55} + y \times \frac{126.9}{143.1} = 8.66 \; \text{g} \).
By solving these equations, we get specific values for \( x \) and \( y \). The simultaneous solving technique involves substitution or elimination methods, ensuring every equation satisfies the given conditions. Practicing such methods augments problem-solving skills involved in chemical calculations.
Molar Mass Calculations
Molar mass is the mass of one mole of a substance, usually expressed in g/mol. It's fundamental in chemistry for conversion between moles and grams. In calculating the masses of CuO and Cu鈧侽, their molar masses are central:
  • Molar mass of CuO is approximately 79.55 g/mol.
  • Molar mass of Cu鈧侽 is approximately 143.1 g/mol.
To find the mass of copper produced when reducing each compound, we set up proportion calculations based on molar masses.
Using these values, we translate amounts of substances from mass to moles, and vice versa, allowing precise calculations for reactions like those in our exercise. This concept is crucial, forming the backbone of stoichiometry鈥攅nsuring chemical equations are balanced both in mass and moles.

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Most popular questions from this chapter

The concentration of a solution of potassium permanganate, \(\mathrm{KMnO}_{4}\), can be determined by titration against a known amount of oxalic acid, \(\mathrm{H}_{2} \mathrm{C}_{2} \mathrm{O}_{4}\), according to the following equation: \(5 \mathrm{H}_{2} \mathrm{C}_{2} \mathrm{O}_{4}(a q)+2 \mathrm{KMnO}_{4}(a q)+3 \mathrm{H}_{2} \mathrm{SO}_{4}(a q) \longrightarrow\) \(10 \mathrm{CO}_{2}(g)+2 \mathrm{MnSO}_{4}(a q)+\mathrm{K}_{2} \mathrm{SO}_{4}(a q)+8 \mathrm{H}_{2} \mathrm{O}(l)\) What is the concentration of a \(\mathrm{KMnO}_{4}\) solution if \(22.35 \mathrm{~mL}\) reacts with \(0.5170 \mathrm{~g}\) of oxalic acid?

Brass is an approximately \(4: 1\) alloy of copper and zinc, along with small amounts of tin, lead, and iron. The mass percents of copper and zinc can be determined by a procedure that begins with dissolving the brass in hot nitric acid. The resulting solution of \(\mathrm{Cu}^{2+}\) and \(\mathrm{Zn}^{2+}\) ions is then treated with aqueous ammonia to lower its acidity, followed by addition of sodium thiocyanate (NaSCN) and sulfurous acid \(\left(\mathrm{H}_{2} \mathrm{SO}_{3}\right)\) to precipitate copper(I) thiocyanate (CuSCN). The solid CuSCN is collected, dissolved in aqueous acid, and treated with potassium iodate \(\left(\mathrm{KIO}_{3}\right)\) to give iodine, which is then titrated with aqueous sodium thiosulfate \(\left(\mathrm{Na}_{2} \mathrm{~S}_{2} \mathrm{O}_{3}\right)\). The filtrate remaining after CuSCN has been removed is neutralized by addition of aqueous ammonia, and a solution of diammonium hydrogen phosphate \(\left(\left(\mathrm{NH}_{4}\right)_{2} \mathrm{HPO}_{4}\right)\) is added to yield a precipitate of zinc ammonium phosphate \(\left(\mathrm{ZnNH}_{4} \mathrm{PO}_{4}\right)\). Heating the precipitate to \(900^{\circ} \mathrm{C}\) converts it to zinc pyrophosphate \(\left(\mathrm{Zn}_{2} \mathrm{P}_{2} \mathrm{O}_{7}\right)\), which is weighed. The equations are (1) \(\mathrm{Cu}(s)+\mathrm{NO}_{3}^{-}(a q) \longrightarrow \mathrm{Cu}^{2+}(a q)+\mathrm{NO}(g)\) (in acid) (2) \(\mathrm{Cu}^{2+}(a q)+\mathrm{SCN}^{-}(a q)+\mathrm{HSO}_{3}^{-}(a q) \longrightarrow\) \(\mathrm{CuSCN}(s)+\mathrm{HSO}_{4}^{-}(a q)\) (in acid) (3) \(\mathrm{Cu}^{+}(a q)+\mathrm{IO}_{3}^{-}(a q) \longrightarrow \mathrm{Cu}^{2+}(a q)+\mathrm{I}_{2}(a q)\) (in acid) (4) \(\mathrm{I}_{2}(a q)+\mathrm{S}_{2} \mathrm{O}_{3}{ }^{2-}(a q) \longrightarrow \mathrm{I}^{-}(a q)+\mathrm{S}_{4} \mathrm{O}_{6}{ }^{2-}(a q)\) (in acid) (5) \(\mathrm{ZnNH}_{4} \mathrm{PO}_{4}(s) \longrightarrow \mathrm{Zn}_{2} \mathrm{P}_{2} \mathrm{O}_{7}(s)+\mathrm{H}_{2} \mathrm{O}(g)+\mathrm{NH}_{3}(g)\) (a) Balance all equations. (b) When a brass sample with a mass of \(0.544 \mathrm{~g}\) was subjected to the preceding analysis, \(10.82 \mathrm{~mL}\) of \(0.1220 \mathrm{M}\) sodium thiosulfate was required for the reaction with iodine. What is the mass percent copper in the brass? (c) The brass sample in part (b) yielded \(0.246 \mathrm{~g}\) of \(\mathrm{Zn}_{2} \mathrm{P}_{2} \mathrm{O}_{7}\). What is the mass percent zinc in the brass?

Write balanced net ionic equations for the following reactions. Note that \(\mathrm{HClO}_{3}\) is a strong acid. (a) \(\mathrm{Fe}(\mathrm{OH})_{3}(s)+\mathrm{H}_{2} \mathrm{SO}_{4}(a q) \rightarrow ?\) (b) \(\mathrm{HClO}_{3}(a q)+\mathrm{NaOH}(a q) \rightarrow ?\)

Iodine, \(\mathrm{I}_{2}\), reacts with aqueous thiosulfate ion in neutral solution according to the balanced equation $$ \mathrm{I}_{2}(a q)+2 \mathrm{~S}_{2} \mathrm{O}_{3}^{2-}(a q) \longrightarrow \mathrm{S}_{4} \mathrm{O}_{6}^{2-}(a q)+2 \mathrm{I}^{-}(a q) $$ How many grams of \(\mathrm{I}_{2}\) are present in a solution if \(35.20 \mathrm{~mL}\) of \(0.150 \mathrm{M} \mathrm{Na}_{2} \mathrm{~S}_{2} \mathrm{O}_{3}\) solution is needed to titrate the \(\mathrm{I}_{2}\) solution?

How would you prepare \(500 \mathrm{~mL}\) of \(0.33 \mathrm{M}\) solution of \(\mathrm{CaCl}_{2}\) from solid \(\mathrm{CaCl}_{2} ?\) Specify the glassware that should be used.
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