How would you prepare \(20.0 \mathrm{~g}\) of iron(III) hydroxide from a \(0.250 \mathrm{M}\) solution of \(\mathrm{Fe}\left(\mathrm{NO}_{3}\right)_{3}\) and a \(0.150 \mathrm{M}\) solution of \(\mathrm{Ba}(\mathrm{OH})_{2} ?\) How much product would you have isolated if your percent yield were \(67.5 \% ?\)

The balanced chemical equation for the reaction between iron(III) nitrate and barium hydroxide is as follows: \[ 2 \mathrm{Fe(NO_3)_3} + 6 \mathrm{Ba(OH)_2} \rightarrow 6 \mathrm{Ba(NO_3)_2} + 2 \mathrm{Fe(OH)_3} \] Step 2: Calculate moles of each reactant

We need to determine the moles of the given reactants by using the molarity and mass formulas: \[M = \frac{mol}{L} \Rightarrow mol = M \times L\] We need to find the volume of \(0.250 \mathrm{M} \ \mathrm{Fe(NO_3)_3}\) and \(0.150 \mathrm{M} \ \mathrm{Ba(OH)_2}\) solutions required to prepare \(20.0 \mathrm{g}\) of \(\mathrm{Fe(OH)_3\). In order to do so, we should first convert mass of iron(III) hydroxide into moles: \[mol = \frac{mass}{molar \ mass} \Rightarrow mol_{Fe(OH)_3} = \frac{20.0 \mathrm{g}}{106.87 \mathrm{g/mol}} = 0.187 \mathrm{mol}\] From the balanced chemical equation, we can see that the mole ratio between \(Fe(NO_3)_3\) and \(Fe(OH)_3\) is 1:1. So, we need 0.187 mol of \(Fe(NO_3)_3\) for the reaction. Similarly, the mole ratio between \(Ba(OH)_2\) and \(Fe(OH)_3\) is 3:1, so we need 3 times more moles of \(Ba(OH)_2\), which is \(3 \times 0.187 \mathrm{mol} = 0.561 \mathrm{mol}\). Now, we can calculate the volume of each solution required: \[ V_{Fe(NO_3)_3} = \frac{0.187 \mathrm{mol}}{0.250 \mathrm{M}} = 0.748 \mathrm{L} \] \[ V_{Ba(OH)_2} = \frac{0.561 \mathrm{mol}}{0.150 \mathrm{M}} = 3.74 \mathrm{L} \] Step 3: Calculate the mass of product

Since the problem asks for the mass of the product when the percent yield is \(67.5\%\), we can calculate the mass of the actual product by multiplying the theoretical mass (20.0 g) with the percent yield and dividing by 100. \[Mass_{actual} = \frac{Percent \ yield \times Theoretical \ mass}{100} \] \[Mass_{actual} = \frac{67.5 \% \times 20.0 \mathrm{g}}{100} = 13.5 \mathrm{g} \] So, the actual mass of iron(III) hydroxide isolated when the percentage yield is 67.5% is 13.5 g.