Chromium has been investigated as a coating for steel cans. The thickness of the chromium film is determined by dissolving a sample of a can in acid and oxidizing the resulting \(\mathrm{Cr}^{3+}\) to $\mathrm{Cr}_{2} \mathrm{O}_{7}^{2-}$ with the peroxydisulfate ion: $$\begin{aligned} \mathrm{S}_{2} \mathrm{O}_{\mathrm{x}}^{2-}(a q)+\mathrm{Cr}^{3+}(a q)+\mathrm{H}_{2} \mathrm{O}(l) & \longrightarrow \mathrm{Cr}_{2} \mathrm{O}_{7}^{2-}(a q) \\ &+\mathrm{SO}_{4}^{2-}(a q)+\mathrm{H}^{+}(a q)(\text { Unbalanced }) \end{aligned} $$ After removal of unreacted \(\mathrm{S}_{2} \mathrm{O}_{8}^{2-},\) an excess of ferrous ammonium sulfate $\left[\mathrm{Fe}\left(\mathrm{NH}_{4}\right)_{2}\left(\mathrm{SO}_{4}\right)_{2} \cdot 6 \mathrm{H}_{2} \mathrm{O}\right]$ is added, reacting with \(\mathrm{Cr}_{2} \mathrm{O}_{7}^{2-}\) produced from the first reaction. The unreacted \(\mathrm{Fe}^{2+}\) from the excess ferrous ammonium sulfate is titrated with a separate \(\mathrm{K}_{2} \mathrm{Cr}_{2} \mathrm{O}_{7}\) solution. The reaction is: $$\begin{array}{c}{\mathrm{H}^{+}(a q)+\mathrm{Fe}^{2+}(a q)+\mathrm{Cr}_{2} \mathrm{O}_{7}^{2-}(a q) \longrightarrow \mathrm{Fe}^{3+}(a q)} \\\ {+\mathrm{Cr}^{3+}(a q)+\mathrm{H}_{2} \mathrm{O}(I) \text { (Unbalanced) }}\end{array}$$ a. Write balanced chemical equations for the two reactions. b. In one analysis, a \(40.0-\mathrm{cm}^{2}\) sample of a chromium-plated can was treated according to this procedure. After dissolution and removal of excess \(\mathrm{S}_{2} \mathrm{O}_{8}^{2}-, 3.000 \mathrm{g}\) of $\operatorname{Fe}\left(\mathrm{NH}_{4}\right)_{2}\left(\mathrm{SO}_{4}\right)_{2} \cdot 6 \mathrm{H}_{2} \mathrm{O}\( was added. It took 8.58 \)\mathrm{mL}$ of 0.0520 \(\mathrm{M} \mathrm{K}_{2} \mathrm{Cr}_{2} \mathrm{O}_{7}\) solution to completely react with the excess \(\mathrm{Fe}^{2+}\) . Calculate the thickness of the chromium film on the can. (The density of chromium is 7.19 $\mathrm{g} / \mathrm{cm}^{3} .$ )

Step by step solution

01

Balancing the chemical equations

For the first reaction, we have to balance the following equation: $$\mathrm{S}_{2}\mathrm{O}_{\mathrm{x}}^{2-}(aq)+\mathrm{Cr}^{3+}(aq)+\mathrm{H}_{2}\mathrm{O}(l) \longrightarrow \mathrm{Cr}_{2}\mathrm{O}_{7}^{2-}(aq)+\mathrm{SO}_{4}^{2-}(aq)+\mathrm{H}^{+}(aq)$$ Balanced equation: $$2\mathrm{S}_{2} \mathrm{O}_{8}^{2-}(aq) + 3\mathrm{Cr}^{3+}(aq) + 8\mathrm{H}_{2}\mathrm{O}(l) \longrightarrow 2\mathrm{Cr}_{2}\mathrm{O}_{7}^{2-}(aq) + 4\mathrm{SO}_{4}^{2-}(aq) + 24\mathrm{H}^{+}(aq)$$ For the second reaction, we have to balance the following equation: $$\mathrm{H}^{+}(aq)+\mathrm{Fe}^{2+}(aq)+\mathrm{Cr}_{2}\mathrm{O}_{7}^{2-}(aq) \longrightarrow \mathrm{Fe}^{3+}(aq)+\mathrm{Cr}^{3+}(aq)+\mathrm{H}_{2}\mathrm{O}(l)$$ Balanced equation: $$14\mathrm{H}^{+}(aq) + 6\mathrm{Fe}^{2+}(aq) + \mathrm{Cr}_{2}\mathrm{O}_{7}^{2-}(aq) \longrightarrow 6\mathrm{Fe}^{3+}(aq) + 2\mathrm{Cr}^{3+}(aq) + 7\mathrm{H}_{2}\mathrm{O}(l)$$

02

Calculate moles of \(\mathrm{Fe}^{2+}\) in the excess ferrous ammonium sulfate

We are given that 3.000 g of ferrous ammonium sulfate was added to the sample. We need to find the number of moles of \(\mathrm{Fe}^{2+}\) present in it. First, let's find the molar mass of the ferrous ammonium sulfate: Molar mass of \(\mathrm{Fe}\left(\mathrm{NH}_{4}\right)_{2}\left(\mathrm{SO}_{4}\right)_{2} \cdot 6\mathrm{H}_{2}\mathrm{O}\) = 1(55.85) + 2(14.01 + 4(1.01)) + 2(32.07 + 4(16.00)) + 12(1.01) + 6(18.02) = 392.14 g/mol Now, we can calculate the moles of ferrous ammonium sulfate: Moles = mass / molar mass = 3.000 g / 392.14 g/mol = 7.654 * 10^{-3} mol Each mole of ferrous ammonium sulfate gives 1 mole of \(\mathrm{Fe}^{2+}\), so there are 7.654 * 10^{-3} mol of \(\mathrm{Fe}^{2+}\) in the excess ferrous ammonium sulfate.

03

Calculate moles of unreacted \(\mathrm{Fe}^{2+}\) and reacted \(\mathrm{Fe}^{2+}\)

It took 8.58 mL of 0.0520 M \(\mathrm{K}_{2}\mathrm{Cr}_{2}\mathrm{O}_{7}\) solution to completely react with the excess \(\mathrm{Fe}^{2+}\). Let's find the moles of \(\mathrm{K}_{2}\mathrm{Cr}_{2}\mathrm{O}_{7}\) used: Moles of \(\mathrm{K}_{2}\mathrm{Cr}_{2}\mathrm{O}_{7}\) = volume * concentration = 8.58 mL * (0.0520 mol/L) * (1 L / 1000 mL) = 4.4616*10^{-4} mol Now, looking at the balanced equation of the second reaction, 1 mole of \(\mathrm{K}_{2}\mathrm{Cr}_{2}\mathrm{O}_{7}\) reacts with 6 moles of \(\mathrm{Fe}^{2+}\). So, we will now find the moles of \(\mathrm{Fe}^{2+}\) that reacted with the \(\mathrm{K}_{2}\mathrm{Cr}_{2}\mathrm{O}_{7}\): Moles of unreacted \(\mathrm{Fe}^{2+}\) = (6 * moles of \(\mathrm{K}_{2}\mathrm{Cr}_{2}\mathrm{O}_{7}\)) = 6 * 4.4616*10^{-4} mol = 2.67696*10^{-3} mol Next, we need to find the moles of \(\mathrm{Fe}^{2+}\) that reacted with \(\mathrm{Cr}_{2}\mathrm{O}_{7}^{2-}\) in the first reaction: Moles of reacted \(\mathrm{Fe}^{2+}\) = moles (total \(\mathrm{Fe}^{2+}\)) - moles (unreacted \(\mathrm{Fe}^{2+}\)) = 7.654 * 10^{-3} mol - 2.67696 * 10^{-3} mol = 4.97704 * 10^{-3} mol

04

Calculate moles of \(\mathrm{Cr}^{3+}\) from the first reaction

We now need to find out the moles of \(\mathrm{Cr}^{3+}\) that must have reacted in the first reaction. From the balanced equation, for every 6 moles of \(\mathrm{Fe}^{2+}\) that reacts, 2 moles of \(\mathrm{Cr}^{3+}\) must have reacted: Moles of \(\mathrm{Cr}^{3+}\) = (2 * moles of reacted \(\mathrm{Fe}^{2+}\)) / 6 = (2 * 4.97704 * 10^{-3}) / 6 = 1.65835 * 10^{-3} mol

05

Calculate mass and volume of chromium

Now, let's calculate the mass of chromium using its molar mass (Molar mass of \(\mathrm{Cr}\) = 51.996 g/mol): Mass of \(\mathrm{Cr}\) = moles of \(\mathrm{Cr}^{3+}\) * molar mass of \(\mathrm{Cr}\) = 1.65835 * 10^{-3} mol * 51.996 g/mol = 0.086156 g Next, let's calculate the volume of chromium using its density (Density of \(\mathrm{Cr}\) = 7.19 g/cm³): Volume of \(\mathrm{Cr}\) = Mass of \(\mathrm{Cr}\) / Density of \(\mathrm{Cr}\) = 0.086156 g / 7.19 g/cm³ = 0.01198 cm³

06

Calculate the thickness of the chromium film

We are given that the area of the sample is 40.0 cm². Now, to find the thickness of the film, we will divide the volume of chromium by the area: Thickness of chromium film = volume / area = 0.01198 cm³ / 40.0 cm² = 2.995 * 10^{-4} cm The thickness of the chromium film on the can is 2.995 * 10^{-4} cm.

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