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Chapter 5: Problem 8

Gold metal reacts with a combination of nitric acid and hydrochloric acid according to this equation: $$ \mathrm{Au}(\mathrm{s})+3 \mathrm{HNO}_{3}(\mathrm{aq})+4 \mathrm{HCl}(\mathrm{aq}) \rightarrow \mathrm{HAuCl}_{4}(\mathrm{aq})+3 \mathrm{NO}_{2}(\mathrm{~g})+3 \mathrm{H}_{2} \mathrm{O}(\ell) $$ a. Verify that this equation is balanced. b. How many Au atoms react with 639 molecules of aqueous \(\mathrm{HNO}_{3}\) ?

Short Answer

Expert verified
The equation is balanced. 213 Au atoms react with 639 molecules of HNO鈧�.

Step by step solution

01

Counting Atoms on the Reactants Side

Begin by counting the number of each type of atom on the reactants' side of the equation. You have one atom of gold (Au), three nitrogen atoms from three molecules of nitric acid (HNO鈧�), nine oxygen atoms from the same nitric acid, four hydrogen atoms from four molecules of hydrochloric acid (HCl), and four chlorine atoms also from HCl.
02

Counting Atoms on the Products Side

Now count the atoms on the products' side of the equation. You have one gold atom in HAuCl鈧�, one nitrogen atom in each NO鈧� molecule resulting in three nitrogen atoms, two oxygen atoms in each NO鈧� summing up to six oxygen atoms, two hydrogen atoms in each molecule of water (H鈧侽) leading to six hydrogen atoms total, and four chlorine atoms from HAuCl鈧�.
03

Matching Atom Counts to Verify Balance

Compare the counts from both sides of the equation. For gold, you have 1 gold atom on each side. For nitrogen, there are 3 atoms on both sides. Oxygen counts match with 9 on the reactants' side (3 NO鈧� groups) and 6 from NO鈧� plus 3 from H鈧侽 on the products' side. For hydrogen, there are 6 atoms each side (from HCl and H鈧侽). Finally, chlorine matches with 4 atoms on each side. All counts match, confirming the equation is balanced.
04

Calculating Mole Ratio to Solve the Question

The balanced equation shows that 1 mol of Au reacts with 3 mol of HNO鈧�. Thus, the ratio of Au atoms to HNO鈧� molecules is 1 to 3.
05

Solving for Gold Atoms Required

According to the ratio, 639 molecules of HNO鈧� correspond to \( \frac{639}{3} \) Au atoms. Calculate \( \frac{639}{3} = 213 \). So, 213 Au atoms react with 639 molecules of HNO鈧�.

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Key Concepts

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

Balancing Equations
Balancing chemical equations is like solving a puzzle. It ensures that matter is conserved in a reaction, meaning the same number of each type of atom appears on both sides of the equation. This concept is crucial because it reflects the law of conservation of mass. In the given equation, the reaction of gold (Au) with nitric and hydrochloric acids, balance is verified by careful atom counting.
  • We start by listing all atoms involved: Au, H, N, O, and Cl.
  • On the reactant side, the equation provides a breakdown: 1 Au, 3 N (from 3 x HNO鈧�), 9 O (from 3 x HNO鈧�), 4 H (from 4 x HCl), and 4 Cl (from 4 x HCl).
  • For the products: 1 Au (in HAuCl鈧�), 3 N (in 3 x NO鈧�), 6 O (2 per NO鈧�) plus 3 O in 3 x H鈧侽, and finally, 6 H (from 3 x H鈧侽), and 4 Cl (in HAuCl鈧�).
We compare counts across the equation: All atoms match up perfectly, confirmed through matching atom counts across Au, N, O, H, and Cl on both sides. This balance demonstrates a correct representation of the chemical reaction.
Atom Counting
Counting atoms in a chemical equation helps to understand how each element is involved and ensures the equation is balanced. It requires methodically breaking down molecular compounds into individual atoms. For a complex reaction, like the one involving gold and acids, it鈥檚 especially vital.
  • Start by identifying each molecule on both reactant and product sides.
  • Break each compound into its atomic components. For example, HNO鈧� splits into H, N, and three O atoms.
  • Repeat for each species, considering stoichiometric coefficients (numbers before molecules) as multipliers for the atom count.
This atom inventory ensures the conservation of elements throughout the reaction. By confirming the equality of atom counts on both sides, we validate that the chemical equation represents the reaction faithfully.
Mole Ratios
Mole ratios are central to solving stoichiometry problems in chemistry. Once a chemical equation is balanced, mole ratios provide a quantitative relationship between reactants and products. For the presented reaction of gold, each mol of gold reacts with 3 mols of \(\text{HNO}_3\).
  • The balanced equation tells us exactly how many moles of one substance are needed to react completely with a certain number of moles of another.
  • In this example, the mole ratio of Au to \(\text{HNO}_3\) is 1:3, derived directly from the balanced equation.
Knowing these ratios allows one to predict how much of each reactant is needed or how much product will be formed. It鈥檚 a powerful tool for calculating material needs in chemistry and related fields.
Stoichiometry
Stoichiometry is the mathematical language of chemical reactions. It involves using mole ratios derived from balanced chemical equations to calculate quantities of reactants or products. In the exercise, we used stoichiometry to determine how many gold atoms react with a given number of \(\text{HNO}_3\) molecules.
  • Begin with the given quantity, here 639 molecules of \(\text{HNO}_3\).
  • Use the established mole ratio: for every 3 molecules of \(\text{HNO}_3\), 1 Au atom reacts.
  • Calculate the result: dividing 639 by 3 yields 213, indicating 213 Au atoms are needed.
Stoichiometry involves calculations that precisely dictate the amounts of reactants consumed and products generated. It's essential for efficiency and accuracy in chemical production and analysis.

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Most popular questions from this chapter

Balance this equation. $$ \mathrm{N}_{2}+\mathrm{H}_{2} \rightarrow \mathrm{NH}_{3} $$

A typical respiratory reaction discussed in the text is the oxidation of glucose \(\left(\mathrm{C}_{6} \mathrm{H}_{12} \mathrm{O}_{6}\right):\) $$ \mathrm{C}_{6} \mathrm{H}_{12} \mathrm{O}_{6}+6 \mathrm{O}_{2} \rightarrow 6 \mathrm{CO}_{2}+6 \mathrm{H}_{2} \mathrm{O} $$ Is this a redox reaction? If so, what are the oxidizing and reducing agents?

Copper metal reacts with nitric acid according to this equation: $$ 3 \mathrm{Cu}(\mathrm{s})+8 \mathrm{HNO}_{3}(\mathrm{aq}) \rightarrow 3 \mathrm{Cu}\left(\mathrm{NO}_{3}\right)_{2}(\mathrm{aq})+2 \mathrm{NO}(\mathrm{g})+4 \mathrm{H}_{2} \mathrm{O}(\ell) $$ a. Verify that this equation is balanced. b. How many Cu atoms will react if 488 molecules of aqueous \(\mathrm{HNO}_{3}\) are reacted?

What alcohol is produced in the reduction of propanal (CH \(_{3} \mathrm{CH}_{2} \mathrm{CHO}\) )?

Which reactions are redox reactions? For those that are redox reactions, identify the oxidizing and reducing agents. a. \(\mathrm{NaOH}+\mathrm{HCl} \rightarrow \mathrm{H}_{2} \mathrm{O}+\mathrm{NaCl}\) b. \(3 \mathrm{Mg}+2 \mathrm{AlCl}_{3} \rightarrow 2 \mathrm{Al}+3 \mathrm{MgCl}_{2}\) c. \(\mathrm{H}_{2} \mathrm{O}_{2}+\mathrm{H}_{2} \rightarrow 2 \mathrm{H}_{2} \mathrm{O}\) \(\mathrm{d} . \mathrm{KCl}+\mathrm{AgNO}_{3} \rightarrow \mathrm{AgCl}+\mathrm{KNO}_{3}\)
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