Predict the outcome of the reactions represented by the following equations by using the activity series, and balance the equations. (a) \(\mathrm{Cu}(s)+\mathrm{HCl}(a q) \longrightarrow\) (b) \(\mathrm{I}_{2}(s)+\operatorname{NaBr}(a q) \longrightarrow\) (c) \(\mathrm{Mg}(s)+\mathrm{CuSO}_{4}(a q) \longrightarrow\) (d) \(\mathrm{Cl}_{2}(g)+\mathrm{KBr}(a q) \longrightarrow\)

Short Answer

Expert verified
(a) No reaction (b) No reaction (c) \( \mathrm{Mg}(s)+\mathrm{CuSO}_{4}(a q) \longrightarrow \mathrm{MgSO}_{4}(a q)+\mathrm{Cu}(s) \) (d) \( \mathrm{Cl}_{2}(g)+2 \mathrm{KBr}(a q) \longrightarrow 2\mathrm{KCl}(a q)+ \mathrm{Br}_{2}(g) \)

Step by step solution

01

Assess Reactions

Using the activity series, determine which reactions will occur. For a reaction to happen, the single element must be higher in the series than the other element it is reacting with in a compound.
02

Predict and Balance Equation (a)

For equation (a), Cu(s) and HCl(aq) are involved. Copper (Cu) is lower in the activity series than Hydrogen (H), so no reaction occurs.
03

Predict and Balance Equation (b)

For equation (b), I2(s) and NaBr(aq) are involved. Iodine (I) is lower in the activity series than Bromine (Br), so no reaction occurs.
04

Predict and Balance Equation (c)

For equation (c), Mg(s) and CuSO4(aq) are involved. Magnesium (Mg) is higher in the activity series than Copper (Cu), so a reaction does occur. The products are MgSO4(aq) and Cu(s). The balanced reaction is: \( \mathrm{Mg}(s)+\mathrm{CuSO}_{4}(a q) \longrightarrow \mathrm{MgSO}_{4}(a q)+\mathrm{Cu}(s) \)
05

Predict and Balance Equation (d)

For equation (d), Cl2(g) and KBr(aq) are involved. Chlorine (Cl) is higher in the activity series than Bromine (Br), so a reaction does occur. The products are KCl and Br2. The balanced reaction is: \( \mathrm{Cl}_{2}(g)+2 \mathrm{KBr}(a q) \longrightarrow 2\mathrm{KCl}(a q)+ \mathrm{Br}_{2}(g) \)

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Key Concepts

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

Predicting Chemical Reactions
Understanding chemical reactions begins with predicting if a reaction will occur. This requires knowledge of the reactivity of the elements involved. The activity series is a crucial tool here; it lists metals in order of decreasing reactivity. A metal higher in the series can displace a metal lower down from its compound in a solution.

For instance, when predicting if a reaction will take place, compare the position of the pure element with the metal in the compound. If the pure metal is higher in the series, a reaction is possible. An example is the reaction between magnesium and copper sulfate. Because magnesium is above copper in the activity series, it can displace copper, leading to a chemical reaction.
Balancing Chemical Equations
After predicting which reactions will occur, the next step is balancing the chemical equations. Balancing ensures that the number of atoms for each element is the same on both sides of the equation, following the law of conservation of mass.

To balance equations, adjust coefficients—the numbers before the species—without changing the subscripts (which would alter the compounds themselves). For example, balancing the reaction between chlorine gas and potassium bromide involves ensuring that there are an equal number of chlorine and bromine atoms on both sides, resulting in the coefficients 2 for KBr and KCl in the balanced equation.
Single Replacement Reactions
Single replacement reactions involve an element reacting with a compound, resulting in the displacement of an atom in that compound. This type of reaction is clearly illustrated in the activity series exercises.

In the context of the provided exercise, when magnesium metal is placed into copper sulfate solution (Mg and CuSO4), magnesium replaces copper due to its higher reactivity, leading to magnesium sulfate and solid copper. Recognizing single replacement reactions aids in predicting product formation and understanding chemical reactivity.
Reactivity of Metals
The reactivity of metals is pivotal for predicting if a single replacement reaction will occur. Reactivity refers to how easily a metal can lose electrons to form positive ions or cations. This propensity is influenced by many factors including atomic structure and the presence of other elements in a reaction.

Metal reactivity is systematically arranged in the activity series, which lists metals from most reactive to least reactive. Elements towards the top, such as lithium and potassium, react vigorously with substances, while those at the bottom, including gold and platinum, are much less reactive. For example, in the exercises provided, magnesium's position above copper in this series indicates that it's more likely to react in the given scenario.

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Most popular questions from this chapter

A \(0.8870-\mathrm{g}\) sample of a mixture of \(\mathrm{NaCl}\) and \(\mathrm{KCl}\) is dissolved in water, and the solution is then treated with an excess of \(\mathrm{AgNO}_{3}\) to yield \(1.913 \mathrm{~g}\) of \(\mathrm{AgCl}\). Calculate the percent by mass of each compound in the mixture.

Write the equation for calculating molarity. Why is molarity a convenient concentration unit in chemistry?

Calculate the volume in milliliters of a \(1.420 \mathrm{M}\) \(\mathrm{NaOH}\) solution required to titrate the following solutions. (a) \(25.00 \mathrm{~mL}\) of a \(2.430 \mathrm{M} \mathrm{HCl}\) solution (b) \(25.00 \mathrm{~mL}\) of a \(4.500 \mathrm{M} \mathrm{H}_{2} \mathrm{SO}_{4}\) solution (c) \(25.00 \mathrm{~mL}\) of a \(1.500 \mathrm{M} \mathrm{H}_{3} \mathrm{PO}_{4}\) solution

(a) Describe a preparation for magnesium hydroxide \(\left[\mathrm{Mg}(\mathrm{OH})_{2}\right]\) and predict its solubility. (b) Milk of magnesia contains mostly \(\mathrm{Mg}(\mathrm{OH})_{2}\) and is effective in treating acid (mostly hydrochloric acid) indigestion. Calculate the volume of a \(0.035 \mathrm{M} \mathrm{HCl}\) solution (a typical acid concentration in an upset stomach) needed to react with two spoonfuls (approximately \(10 \mathrm{~mL}\) ) of milk of magnesia [at \(0.080 \mathrm{~g}\) \(\left.\mathrm{Mg}(\mathrm{OH})_{2} / \mathrm{mL}\right]\).

The molar mass of a certain metal carbonate, \(\mathrm{MCO}_{3}\), can be determined by adding an excess of \(\mathrm{HCl}\) acid to react with all the carbonate and then "back titrating" the remaining acid with a \(\mathrm{NaOH}\) solution. (a) Write an equation for these reactions. (b) In a certain experiment, \(20.00 \mathrm{~mL}\) of \(0.0800 \mathrm{M} \mathrm{HCl}\) were added to a 0.1022 -g sample of \(\mathrm{MCO}_{3}\). The excess HCl required \(5.64 \mathrm{~mL}\) of \(0.1000 \mathrm{M} \mathrm{NaOH}\) for neutralization. Calculate the molar mass of the carbonate and identify \(\mathrm{M}\).

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