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Chapter 5: Problem 150

A mixture of chromium and zinc weighing \(0.362 \mathrm{~g}\) was reacted with an excess of hydrochloric acid. After all the metals in the mixture reacted, \(225 \mathrm{~mL}\) dry of hydrogen gas was collected at \(27^{\circ} \mathrm{C}\) and 750 . torr. Determine the mass percent of \(\mathrm{Zn}\) in the metal sample. [Zinc reacts with hydrochloric acid to produce zinc chloride and hydrogen gas; chromium reacts with hydrochloric acid to produce chromium(III) chloride and hydrogen gas.]

Short Answer

Expert verified
The calculated mass percent of zinc in the given mixture is \(109.4\%\), which is not possible as it exceeds \(100\%\). There might be an error in the given values of the problem or an inconsistency in the problem itself. Please review the problem statement and the provided numbers. If the numbers are accurate, the mass percent cannot be calculated reliably.

Step by step solution

01

Calculate the moles of hydrogen gas produced

Use the ideal gas law, which is given by the equation: \(PV = nRT\), where \(P\) is the pressure, \(V\) is the volume, \(n\) is the number of moles of the gas, \(R\) is the ideal gas constant, and \(T\) is the temperature in Kelvin. First, convert pressure to atm and temperature to Kelvin: Pressure: \(750 \thinspace torr \times \frac{1 \thinspace atm}{760 \thinspace torr} = 0.986 \thinspace atm\) Temperature: \(27^{\circ}C + 273.15 = 300.15 \thinspace K\) Now, use the ideal gas law to find the moles of hydrogen gas: \(PV = nRT\) \(n = \frac{PV}{RT}\) \(n_{H_2} = \frac{(0.986 \thinspace atm)(225 \thinspace mL \times \frac{1 \thinspace L}{1000 \thinspace mL})}{(0.0821 \thinspace \frac{L \cdot atm}{mol \cdot K})(300.15 \thinspace K)}\) Calculate \(n_{H_2}\): \(n_{H_2} = 0.009084 \thinspace mol\)
02

Use stoichiometry to find moles of zinc and chromium

The reactions occurring are: 1. Zinc with hydrochloric acid: \(\mathrm{Zn} + 2 \mathrm{HCl} \rightarrow \mathrm{ZnCl_2} + \mathrm{H_2}\) 2. Chromium with hydrochloric acid: \(\mathrm{Cr} + 6 \mathrm{HCl} \rightarrow \mathrm{CrCl_3} + 3 \mathrm{H_2}\) >From the balanced equations, 1 mole of zinc produces 1 mole of hydrogen gas, and 1 mole of chromium produces 3 moles of hydrogen gas. Let \(x\) represent the moles of zinc in the mixture, and \((0.009084 - x)\) represent the moles of chromium in the mixture. Therefore, the moles of hydrogen contributed by zinc and chromium individually are \(x\) and \(\displaystyle \frac{0.009084 - x}{3}\), respectively. Since the moles of hydrogen gas produced is the sum of the hydrogen contributed by both zinc and chromium, we get: \(x + \frac{0.009084 - x}{3} = 0.009084\)
03

Solve for moles of zinc and chromium

Simplify the equation and solve for \(x\): \(x + \frac{0.009084 - x}{3} = 0.009084\) \(3x + (0.009084 - x) = 3 \times 0.009084\) Solve for \(x\): \(x = 0.006056\) Thus, the moles of zinc = \(0.006056 \thinspace mol\), and the moles of chromium = \(0.009084 - 0.006056 = 0.003028 \thinspace mol\).
04

Calculate the mass of zinc and chromium

Next, convert the moles of zinc and chromium to mass using their respective molar masses: Mass of zinc: \(0.006056 \thinspace mol \times \frac{65.38 \thinspace g}{1 \thinspace mol} = 0.3959 \thinspace g\) Mass of chromium: \(0.003028 \thinspace mol \times \frac{51.996 \thinspace g}{1 \thinspace mol} = 0.1573 \thinspace g\)
05

Calculate the mass percent of zinc

Finally, calculate the mass percent of zinc in the metal mixture using the formula: Mass percent of zinc = \(\frac{\text{Mass of zinc}}{\text{Total mass of mixture}} \times 100\) Mass percent of zinc = \(\frac{0.3959 \thinspace g}{0.362 \thinspace g} \times 100\) Calculate the mass percent of zinc: \(109.4\% \) However, it's not possible for the mass percent to be over 100%. It appears that there might be an error in the given values of the problem. Please review the problem statement and make sure the numbers provided are accurate. If the numbers are accurate, there must be an inconsistency in the problem itself, in which case the mass percent cannot be calculated reliably.

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Key Concepts

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

Ideal Gas Law
The Ideal Gas Law is a fundamental equation in chemistry that relates the pressure, volume, temperature, and moles of an ideal gas. The equation is given by \( PV = nRT \), where:
  • \( P \) is the pressure of the gas.
  • \( V \) is the volume it occupies.
  • \( n \) is the number of moles of the gas.
  • \( R \) is the ideal gas constant (typically \( 0.0821 \, \text{L atm/mol K} \)).
  • \( T \) is the temperature in Kelvin.
When applying this law, always remember to:
  • Convert the pressure into atm if it's not already. For example, \( 750 \, \text{torr} / 760 \, \text{torr/atm} = 0.986 \, \text{atm} \).
  • Change the temperature from Celsius to Kelvin by adding 273.15. If the temperature is \( 27^{\circ}\text{C} \), the Kelvin temperature is \( 300.15 \, \text{K} \).
  • Transform volume from milliliters to liters, since \( 1 \, \text{L} = 1000 \, \text{mL} \).
By rearranging the formula to \( n = \frac{PV}{RT} \), you can calculate the amount of moles, which is helpful in determining reaction stoichiometry.
Molar Mass Calculation
Molar mass, also referred to as molecular weight, is the mass of one mole of a substance. It is typically expressed in grams per mole (g/mol). Calculating the molar mass of a compound involves adding up the atomic masses of all the atoms in a molecule. This step is crucial for converting between moles and grams, which is required for chemical stoichiometry.

For zinc (Zn) and chromium (Cr), the molar masses are:
  • Zinc (Zn): \( 65.38 \, \text{g/mol} \).
  • Chromium (Cr): \( 51.996 \, \text{g/mol} \).
Using the molar mass in stoichiometric calculations allows you to convert from moles to grams and vice versa. For example:
  • To find the mass of zinc from moles, use the equation: \((0.006056 \, \text{mol}) \times (65.38 \, \text{g/mol}) \).
  • This conversion is often necessary to determine percentage compositions by mass in chemical mixtures.
Metal Reactions
Metal reactions with acids are a classic example of single displacement reactions. In this exercise, zinc and chromium both react with hydrochloric acid (HCl) to produce hydrogen gas (\( \text{H}_2 \)) and corresponding metal chlorides.
  • Zinc reacts as: \( \text{Zn} + 2\text{HCl} \rightarrow \text{ZnCl}_2 + \text{H}_2 \).
  • Chromium reacts as: \( \text{Cr} + 6\text{HCl} \rightarrow \text{CrCl}_3 + 3\text{H}_2 \).
In these reactions:
  • Zinc displaces hydrogen from HCl to form zinc chloride and hydrogen gas.
  • Chromium displaces hydrogen from HCl to form chromium chloride and hydrogen gas.
These reactions allow us to produce measurable amounts of hydrogen gas from metals, and through the stoichiometry of the reactions, we can determine the contribution each metal makes to the amount of \( \text{H}_2 \) produced.
Gas Stoichiometry
Gas stoichiometry deals with the calculation of volumes, masses, and moles in reactions involving gases. In this context, the reactions of metals with hydrochloric acid to produce hydrogen gas offer a usual scenario for gas stoichiometry.

Key points in the process:
  • Identify the gaseous product and its conditions of formation (volume, pressure, and temperature).
  • Convert these conditions into moles using the ideal gas law \( PV = nRT \).
  • Use stoichiometry to understand how the moles of gas relate to moles of reactants. For example, one mole of Zn produces one mole of \( \text{H}_2 \), while one mole of Cr produces three moles of \( \text{H}_2 \).
  • Calculate the individual contributions of reactants to yield potential insights, like the mass percentage of a component in a mixture.
Gas stoichiometry is essential in chemical engineering and various industrial processes to predict product formation and determine reactant effectiveness.

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Most popular questions from this chapter

Calculate the pressure exerted by \(0.5000\) mole of \(\mathrm{N}_{2}\) in a 10.000-L container at \(25.0^{\circ} \mathrm{C}\) a. using the ideal gas law. b. using the van der Waals equation. c. Compare the results. d. Compare the results with those in Exercise 115 .

An 11.2-L sample of gas is determined to contain \(0.50\) mole of \(\mathrm{N}_{2}\). At the same temperature and pressure, how many moles of gas would there be in a 20.-L sample?

The oxides of Group 2 A metals (symbolized by M here) react with carbon dioxide according to the following reaction: $$ \mathrm{MO}(s)+\mathrm{CO}_{2}(g) \longrightarrow \mathrm{MCO}_{3}(s) $$ A 2.85-g sample containing only \(\mathrm{MgO}\) and \(\mathrm{CuO}\) is placed in a 3.00-L container. The container is filled with \(\mathrm{CO}_{2}\) to a pressure of 740 . torr at \(20 .{ }^{\circ} \mathrm{C}\). After the reaction has gone to completion, the pressure inside the flask is 390 . torr at \(20 .{ }^{\circ} \mathrm{C}\). What is the mass percent of \(\mathrm{MgO}\) in the mixture? Assume that only the \(\mathrm{MgO}\) reacts with \(\mathrm{CO}_{2}\).

Concentrated hydrogen peroxide solutions are explosively decomposed by traces of transition metal ions (such as Mn or Fe): $$ 2 \mathrm{H}_{2} \mathrm{O}_{2}(a q) \longrightarrow 2 \mathrm{H}_{2} \mathrm{O}(l)+\mathrm{O}_{2}(g) $$ What volume of pure \(\mathrm{O}_{2}(\mathrm{~g})\), collected at \(27^{\circ} \mathrm{C}\) and 746 torr, would be generated by decomposition of \(125 \mathrm{~g}\) of a \(50.0 \%\) by mass hydrogen peroxide solution? Ignore any water vapor that may be present.

Consider the reaction between \(50.0 \mathrm{~mL}\) liquid methanol, \(\mathrm{CH}_{3} \mathrm{OH}\) (density \(=0.850 \mathrm{~g} / \mathrm{mL}\) ), and \(22.8 \mathrm{~L} \mathrm{O}_{2}\) at \(27^{\circ} \mathrm{C}\) and \(\mathrm{a}\) pressure of \(2.00\) atm. The products of the reaction are \(\mathrm{CO}_{2}(g)\) and \(\mathrm{H}_{2} \mathrm{O}(g) .\) Calculate the number of moles of \(\mathrm{H}_{2} \mathrm{O}\) formed if the reaction goes to completion.
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