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Chapter 6: Problem 58

Although Cu is not sufficiently active to react with acids, it can be dissolved by concentrated nitric acid, which functions as an oxidizing agent according to the following equation: \(\mathrm{Cu}(s)+4 \mathrm{HNO}_{3}(a q) \longrightarrow\) $$\mathrm{Cu}\left(\mathrm{NO}_{3}\right)_{2}(a q)+2 \mathrm{NO}_{2}(g)+2 \mathrm{H}_{2} \mathrm{O}(l)$$ (a) Write the net ionic equation for this process. (b) Is \(35.0 \mathrm{~g}\) of \(\mathrm{HNO}_{3}\) sufficient to dissolve \(5.00 \mathrm{~g}\) of copper?

Short Answer

Expert verified
35.0 g of \( \mathrm{HNO}_3 \) is sufficient to dissolve 5.00 g of \( \mathrm{Cu} \).

Step by step solution

01

Write the Balanced Net Ionic Equation

To write the net ionic equation, first identify the ions involved in the reaction. The complete reaction is: \[ \mathrm{Cu}(s) + 4 \mathrm{HNO}_3(aq) \longrightarrow \mathrm{Cu}(\mathrm{NO}_3)_2(aq) + 2 \mathrm{NO}_2(g) + 2 \mathrm{H}_2\mathrm{O}(l) \] \( \mathrm{Cu} \) and \( \mathrm{NO}_3^- \) ions are involved. Remove the spectator ions to get the net ionic equation: \[ \mathrm{Cu}(s) + 4 \mathrm{H^+}(aq) + 2 \mathrm{NO}_3^-(aq) \longrightarrow \mathrm{Cu}^{2+}(aq) + 2 \mathrm{NO}_2(g) + 2 \mathrm{H}_2\mathrm{O}(l) \] This is the balanced net ionic equation, with the elimination of the spectator ions (e.g., complexed \( \mathrm{NO}_3^- \) in the products).
02

Calculate the Moles of \( ext{HNO}_3 ext{ Required} o ext{Determine Moles of Cu that can be Dissolved} ",

Calculate the moles of \( \mathrm{Cu} \) using its molar mass: \[ 5.00 \text{ g Cu} \times \frac{1 \text{ mol Cu}}{63.55 \text{ g Cu}} = 0.0787 \text{ mol Cu} \] \( 4 \text{ mol of } \mathrm{HNO}_3 \) is needed to dissolve 1 mol of \( \mathrm{Cu} \). Therefore, the moles of \( \mathrm{HNO}_3 \) required for \( 0.0787 \text{ mol Cu} \) is: \[ 0.0787 \text{ mol Cu} \times \frac{4 \text{ mol } \mathrm{HNO}_3}{1 \text{ mol Cu}} = 0.3148 \text{ mol } \mathrm{HNO}_3 \]
03

Calculate the Mass of \\ \\ \( ext{HNO}_3 ext{ Required}\)

Convert the moles of \( \mathrm{HNO}_3 \) needed to mass using its molar mass (63.01 g/mol): \[ 0.3148 \text{ mol } \mathrm{HNO}_3 \times \frac{63.01 \text{ g } \mathrm{HNO}_3}{1 \text{ mol } \mathrm{HNO}_3} = 19.82 \text{ g } \mathrm{HNO}_3 \]
04

Compare with the Available Amount of \\\\\\ \( ext{HNO}_3 ext{ Given}\)

The problem states that we have 35.0 g of \( \mathrm{HNO}_3 \). Since 19.82 g is required to dissolve 5.00 g of \( \mathrm{Cu} \), 35.0 g of \( \mathrm{HNO}_3 \) is more than sufficient to dissolve the copper.

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Key Concepts

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

Chemical Reaction
A chemical reaction involves the transformation of reactants into products. In this case, copper (Cu) reacts with concentrated nitric acid (HNO鈧�) to form copper(II) nitrate, nitrogen dioxide (NO鈧�), and water.
  • The reactants are Cu and HNO鈧�.
  • The products are Cu(NO鈧�)鈧�, NO鈧�, and H鈧侽.
Chemical equations must be balanced to ensure the conservation of mass. This means the number of atoms for each element must be the same on both sides of the reaction. Balancing helps us understand the stoichiometry of the reaction and what amount of each reactant is necessary.
Molar Mass Calculation
Molar mass is the mass of one mole of a substance and is expressed in g/mol. It is a crucial component in stoichiometry, allowing us to convert between grams and moles.
For copper, the molar mass is 63.55 g/mol. To find the number of moles in a sample, divide the mass of the sample by its molar mass:
\[ \text{Moles of Cu} = \frac{5.00 \text{ g Cu}}{63.55 \text{ g/mol}} = 0.0787 \text{ mol Cu} \]
Similarly, you can calculate the moles of HNO鈧� needed by using its molar mass, 63.01 g/mol. This calculation helps determine how much of the reactant is needed to complete a reaction.
Oxidizing Agent
In a redox reaction, an oxidizing agent is the substance that gains electrons and is reduced.
  • In this reaction, HNO鈧� acts as the oxidizing agent.
  • It causes copper to oxidize by accepting electrons from it.
As copper loses electrons and forms Cu虏鈦�, nitric acid is reduced to nitrogen dioxide (NO鈧�). Understanding oxidizing and reducing agents is essential in predicting how a reaction will proceed and what products will form.
Stoichiometry
Stoichiometry involves the calculation of reactants and products in chemical reactions. It uses the coefficients in a balanced chemical equation to determine the proportional relationships between different substances.
In this particular exercise, stoichiometry helps us find out how much HNO鈧� is needed to dissolve a given amount of Cu.
  • 4 moles of HNO鈧� are required to dissolve 1 mole of Cu.
  • Calculations showed that 0.3148 moles of HNO鈧� are needed for 0.0787 moles of Cu.
Applying stoichiometry ensures that we use the exact proportions of each reactant, avoiding waste and ensuring completeness of the reaction.

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Most popular questions from this chapter

Batrachotoxin, \(\mathrm{C}_{31} \mathrm{H}_{42} \mathrm{~N}_{2} \mathrm{O}_{6},\) an active component of South American arrow poison, is so toxic that \(0.05 \mu \mathrm{g}\) can kill a person. How many molecules is this?

Magnesium metal burns in oxygen to form magnesium oxide, \(\mathrm{MgO}\) (a) Write a balanced equation for the reaction. (b) How many grams of oxygen are needed to react with \(25.0 \mathrm{~g}\) of \(\mathrm{Mg}\) ? How many grams of \(\mathrm{Mg} \mathrm{O}\) will result? (c) How many grams of \(\mathrm{Mg}\) are needed to react with \(25.0 \mathrm{~g}\) of \(\mathrm{O}_{2}\) ? How many grams of \(\mathrm{MgO}\) will result?

One method for preparing pure iron from \(\mathrm{Fe}_{2} \mathrm{O}_{3}\) is by reaction with carbon monoxide. \(\mathrm{Fe}_{2} \mathrm{O}_{3}(s)+\mathrm{CO}(g) \longrightarrow \mathrm{Fe}(s)+\mathrm{CO}_{2}(g)\) (a) Balance the equation. (b) How many grams of \(\mathrm{CO}\) are needed to react with \(3.02 \mathrm{~g}\) of \(\mathrm{Fe}_{2} \mathrm{O}_{3}\) ? (c) How many grams of \(\mathrm{CO}\) are needed to react with \(1.68 \mathrm{~mol}\) of \(\mathrm{Fe}_{2} \mathrm{O}_{3} ?\)

Ethyl acetate reacts with \(\mathrm{H}_{2}\) in the presence of a catalyst to yield ethanol. \(\mathrm{C}_{4} \mathrm{H}_{8} \mathrm{O}_{2}(l)+\mathrm{H}_{2}(g) \longrightarrow \mathrm{C}_{2} \mathrm{H}_{6} \mathrm{O}(l)\) (a) Write a balanced equation for the reaction. (b) How many moles of ethanol are produced by reaction of \(1.5 \mathrm{~mol}\) of ethyl acetate? (c) How many grams of ethanol are produced by reaction of \(1.5 \mathrm{~mol}\) of ethyl acetate with \(\mathrm{H}_{2} ?\) (d) How many grams of ethanol are produced by reaction of \(12.0 \mathrm{~g}\) of ethyl acetate with \(\mathrm{H}_{2} ?\) (e) How many grams of \(\mathrm{H}_{2}\) are needed to react with \(12.0 \mathrm{~g}\) of ethyl acetate?

The active ingredient in milk of magnesia (an antacid) is magnesium hydroxide, \(\mathrm{Mg}(\mathrm{OH})_{2}\). A typical dose (one tablespoon) contains \(1.2 \mathrm{~g}\) of \(\mathrm{Mg}(\mathrm{OH})_{2} .\) Calculate (a) the molar mass of magnesium hydroxide and (b) the amount of magnesium hydroxide (in moles) in one tablespoon.
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