Question

1. calculate the theoretical yield in grams for this reaction. assume that aluminum is the limiting reagent and use the molar mass of al, the reaction stoichiometry, and the molar mass of alum in your calculation. an example calculation is shown in box 2 on page 4. use element molar masses to the nearest 0.01 g/mol in your calculations.

0.5000g al × \\(\frac{1\text{ mol al}}{26.98\text{ g al}}\\) × \\(\frac{2\text{ mol alum}}{2\text{ mol al}}\\) × \\(\frac{474.4\text{ g}}{1\text{ mol alum}}\\) =

Explanation:

Step 1: Analyze the given calculation setup

We have the mass of Al as \( 0.5000\,\text{g} \), molar mass of Al is \( 26.98\,\text{g/mol} \), the stoichiometric ratio from the reaction (assuming the reaction forms alum, and the ratio of Al to alum is \( \frac{2\,\text{mol Al}}{2\,\text{mol alum}} \) or simplified \( \frac{1\,\text{mol Al}}{1\,\text{mol alum}} \)? Wait, looking at the fractions: \( \frac{1\,\text{mol Al}}{26.98\,\text{g Al}} \), then \( \frac{2\,\text{mol alum}}{2\,\text{mol Al}} \) (which simplifies to \( \frac{1\,\text{mol alum}}{1\,\text{mol Al}} \)), and then molar mass of alum is \( 474.4\,\text{g/mol} \). Wait, let's do the calculation step by step.

First, convert mass of Al to moles of Al:
Moles of Al \( = \frac{\text{mass of Al}}{\text{molar mass of Al}} = \frac{0.5000\,\text{g}}{26.98\,\text{g/mol}} \)

Step 2: Apply stoichiometric ratio

From the reaction, the ratio of moles of alum to moles of Al: looking at the fraction \( \frac{2\,\text{mol alum}}{2\,\text{mol Al}} = 1 \) (so moles of alum = moles of Al)

Step 3: Convert moles of alum to mass of alum

Mass of alum \( = \) moles of alum \( \times \) molar mass of alum
Moles of Al \( = \frac{0.5000}{26.98} \approx 0.01853\,\text{mol} \)
Moles of alum \( = 0.01853\,\text{mol} \) (since ratio is 1)
Mass of alum \( = 0.01853\,\text{mol} \times 474.4\,\text{g/mol} \)

Now let's compute the entire expression:
\( 0.5000\,\text{g Al} \times \frac{1\,\text{mol Al}}{26.98\,\text{g Al}} \times \frac{2\,\text{mol alum}}{2\,\text{mol Al}} \times \frac{474.4\,\text{g alum}}{1\,\text{mol alum}} \)

Simplify the ratios: \( \frac{2}{2} = 1 \), so it becomes \( 0.5000 \times \frac{1}{26.98} \times 1 \times 474.4 \)

Calculate \( \frac{0.5000 \times 474.4}{26.98} \)

First, \( 0.5000 \times 474.4 = 237.2 \)

Then, \( \frac{237.2}{26.98} \approx 8.79 \) (rounded to two decimal places)

Wait, let's do the calculation more accurately:

\( 0.5000 \div 26.98 = 0.018532246 \)

\( 0.018532246 \times 474.4 = 0.018532246 \times 474.4 \)

Calculate \( 0.018532246 \times 474.4 \):

\( 0.018532246 \times 400 = 7.4128984 \)

\( 0.018532246 \times 70 = 1.29725722 \)

\( 0.018532246 \times 4.4 = 0.0815418824 \)

Sum: \( 7.4128984 + 1.29725722 = 8.71015562 + 0.0815418824 = 8.7916975 \approx 8.79\,\text{g} \)

Answer:

The theoretical yield of alum is approximately \( \boldsymbol{8.79\,\text{g}} \) (depending on the exact molar masses used, but following the given values, this is the result).