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Chapter 4: Problem 142

Zinc and magnesium metal each react with hydrochloric acid according to the following equations: $$\mathrm{Zn}(s)+2 \mathrm{HCl}(a q) \longrightarrow \mathrm{ZnCl}_{2}(a q)+\mathrm{H}_{2}(g)$$ $$\mathrm{Mg}(s)+2 \mathrm{HCl}(a q) \longrightarrow \mathrm{MgCl}_{2}(a q)+\mathrm{H}_{2}(g)$$ A 10.00-g mixture of zinc and magnesium is reacted with the stoichiometric amount of hydrochloric acid. The reaction mixture is then reacted with 156 mL of 3.00 M silver nitrate to produce the maximum possible amount of silver chloride. a. Determine the percent magnesium by mass in the original mixture. b. If 78.0 mL of HCl was added, what was the concentration of the HCl?

Short Answer

Expert verified
a. The mass percentage of magnesium in the original mixture is 20.58%. b. The concentration of the hydrochloric acid added to the mixture is 6.00 M.

Step by step solution

01

Moles of AgNO3

\(moles_{AgNO3} = (concentration_{AgNO3})(volume_{AgNO3})\) To find moles of AgNO3, we need to multiply the given concentration (3.00 M) with the given volume (156 mL). \(moles_{AgNO3} = 3.00\,M \times 0.156\,L = 0.468\,mol\) Since AgNO3 reacts with chloride ions (from ZnCl2 and MgCl2) in a 1:1 ratio, the amount of chloride ions is equal to the amount of AgNO3.
02

Moles of chloride ions

\(moles_{Cl^-} = moles_{AgNO3} = 0.468\,mol\) **Step 2: Obtain the equations and combine them** Now, let's put the zinc and magnesium reactions equations into the form of moles.
03

Equations

\(\frac{1}{2} moles_{Cl^-} = moles_{Zn}\) (from Zn reaction) \(\frac{1}{2} moles_{Cl^-} = moles_{Mg}\) (from Mg reaction) Adding both equations:
04

Combined equation

\(\frac{1}{2} moles_{Cl^-} = moles_{Zn} + moles_{Mg}\) **Step 3: Solve the combined equation** Now, let's solve the combined equation to find moles of Zn and Mg.
05

Calculate moles of Zn and Mg

\(\frac{1}{2} \times 0.468\, mol = moles_{Zn} + moles_{Mg}\) \(0.234\,mol = moles_{Zn} + moles_{Mg}\) **Step 4: Find the mass of Zn and Mg** Since we found the moles of Zn and Mg combined (0.234 mol), we can calculate their individual masses, which must sum up to the total mass of the mixture (10.00 g). Let mass of Zn = x grams mass of Mg = (10 - x) grams We will now transform the moles equation using the molar mass of Zn (65.38 g/mol) and Mg (24.305 g/mol).
06

Mass equation

\(\frac{x}{65.38} + \frac{10-x}{24.305} = 0.234\) **Step 5: Solve the mass equation and find mass percentage of Mg** Now, we need to solve the equation for x (= mass of Zn) and find the mass of magnesium.
07

Solving Equation

Using a calculator, we find that \(x = 7.942\,g\). So, mass of Mg = 10 - 7.942 = 2.058 g Now, let's find the mass percentage of magnesium (Mg) in the mixture:
08

Mass percentage of Mg

\(MassPercentage_{Mg} = \frac{mass_{Mg}}{(mass_{Mg} + mass_{Zn})} \times 100\%\) \(MassPercentage_{Mg} = \frac{2.058\,g}{(10.00\,g)} \times 100\% = 20.58\%\), a. Thus, the mass percentage of magnesium in the original mixture is 20.58%. **Step 6: Find the concentration of HCl** Finally, we find the concentration of HCl. We know 78.0 mL of HCl was added, and the stoichiometric amount was used. Since both Zn and Mg have reacted with 2 moles of HCl for every mole of the metal, the moles of HCl that reacted with both metals will be twice the total moles of Zn and Mg (0.234 mol).
09

Moles of HCl

\(moles_{HCl} = 2 \times 0.234\,mol = 0.468\,mol\) Now, we can calculate the concentration of HCl:
10

HCl Concentration

\(concentration_{HCl} = \frac{moles_{HCl}}{volume_{HCl}}\) \(concentration_{HCl} = \frac{0.468\,mol}{0.078\,L} = 6.00\,M\) b. The concentration of the hydrochloric acid added to the mixture is 6.00 M.

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