Assuming you start with 1.25 g of pure aluminum, calculate the following:

1) The amount of potassium hydroxide, $KOH$, in grams, needed to react with all of the $1.25\text{ g}$ of $Al$. The reaction is:

$2 Al(s) + 2 KOH(aq) + 6 H_2O(l) \longrightarrow 2 K^{+}(aq) + 2 Al(OH)_4^{-}(aq) + 3 H_2(g)$

2) The amount of $KOH$ needed, in grams, in reaction 1 if you want to have $10\%$ excess $KOH$.

3) The amount of water, in milliliters, that will react with the $Al$ and $KOH$ in reaction 1.

4) The amount, in milliliters, of $9\text{ M}$ sulfuric acid needed for reaction 2:

$2 Al(OH)_4^{-}(aq) + H_2SO_4(aq) \longrightarrow 2 Al(OH)_3(s) + SO_4^{2-}(aq) + 2 H_2O(l)$

To find the required amounts of chemical reagents, we first determine the moles of aluminum ($Al$) we are starting with. Using the molar mass of aluminum ($26.98\text{ g/mol}$):
\[n(\text{Al}) = \frac{1.25\text{ g}}{26.98\text{ g/mol}} \approx 0.04633\text{ mol}\]
Next, we analyze the stoichiometry of the first reaction:
\[2\text{Al}(s) + 2\text{KOH}(aq) + 6\text{H}_2\text{O}(l) \longrightarrow 2\text{K}^{+}(aq) + 2\text{Al}(\text{OH})_4^{-}(aq) + 3\text{H}_2(g)\]
According to the balanced equation, 2 moles of $Al$ react with 2 moles of $KOH$. Therefore, the moles of $KOH$ required are equal to the moles of $Al$, which is $0.04633\text{ mol}$. The mass of $KOH$ needed (molar mass $\approx 56.11\text{ g/mol}$) is:
\[\text{mass(KOH)} = 0.04633\text{ mol} \times 56.11\text{ g/mol} \approx 2.60\text{ g}\]

Calculating KOH with 10% excess

If we want a $10\%$ excess of $KOH$, we first find the base mass required, which we calculated as $2.60\text{ g}$, and then multiply by $1.10$ to account for the excess:
\[\text{mass(KOH with excess)} = 2.60\text{ g} \times 1.10 \approx 2.86\text{ g}\]
This ensures that there is more than enough reagent to fully react with the available aluminum.

Calculating Water required for the reaction

For the volume of water, the reaction stoichiometry indicates that 2 moles of $Al$ require 6 moles of $H_2O$. This means 1 mole of $Al$ requires 3 moles of $H_2O$. Thus, the moles of $H_2O$ needed are:
\[n(\text{H}_2\text{O}) = 3 \times n(\text{Al}) = 3 \times 0.04633\text{ mol} \approx 0.13899\text{ mol}\]
Using the molar mass of water ($18.02\text{ g/mol}$), the mass of $H_2O$ is:
\[\text{mass(H}_2\text{O}) = 0.13899\text{ mol} \times 18.02\text{ g/mol} \approx 2.50\text{ g}\]
Assuming the density of water is $1.00\text{ g/mL}$, the volume required is $2.50\text{ mL}$.

Calculating H2SO4 volume for neutralization

The second reaction is:
\[2\text{Al}(\text{OH})_4^{-}(aq) + \text{H}_2\text{SO}_4(aq) \longrightarrow 2\text{Al}(\text{OH})_3(s) + \text{SO}_4^{2-}(aq) + 2\text{H}_2\text{O}(l)\]
The moles of $Al(OH)_4^{-}$ produced in the first reaction are equal to the initial moles of $Al$, which is $0.04633\text{ mol}$. The balanced second reaction shows that 2 moles of $Al(OH)_4^{-}$ react with 1 mole of $H_2SO_4$. Therefore, the moles of $H_2SO_4$ needed are:
\[n(\text{H}_2\text{SO}_4) = \frac{n(\text{Al})}{2} = \frac{0.04633\text{ mol}}{2} \approx 0.023165\text{ mol}\]
Given the concentration of the sulfuric acid is $9\text{ M}$, the volume in liters is:
\[V(\text{H}_2\text{SO}_4) = \frac{0.023165\text{ mol}}{9\text{ mol/L}} \approx 0.002574\text{ L}\]
Converting this volume to milliliters, we get:
\[V(\text{H}_2\text{SO}_4) = 0.002574\text{ L} \times 1000\text{ mL/L} \approx 2.57\text{ mL}\]

Examples

Understanding chemical stoichiometry is essential in laboratory settings and industrial manufacturing. For instance, when producing aluminum hydroxide for use in water purification or as an antacid, chemists must calculate the exact quantities of reagents to minimize waste and ensure high purity. By calculating the required masses and volumes of precursors like KOH and H2SO4, they can control the final product quality while keeping costs low and following safety protocols. This application ensures that industrial processes remain efficient and sustainable.