Question

Type the correct answer in the box. Given: 4Al 3O2 → 2Al2O3 In this chemical reaction, how many grams of Al2O3 will be produced if 375 grams of Al react completely? Express your answer to three significant figures. There will be grams of Al2O3.

Answer 1

To find the mass of Al2O3 formed from 375 g of Al, one must use stoichiometry to convert grams of Al to moles, use the balanced equation to find moles of Al2O3, and then convert moles of Al2O3 back to grams. The result is 708 g of Al2O3.

Step-by-step explanation:

To determine the mass of Al2O3 produced from 375 grams of Al using the given chemical reaction 4Al + 3O2 → 2Al2O3, we first need to calculate the molar mass of aluminium (Al), which is approximately 26.98 g/mol. Then, we'll find the molar mass of aluminium oxide (Al2O3), which is 101.96 g/mol (26.98 g/mol × 2 + 16.00 g/mol × 3). By using stoichiometry, we can find the relationship between the moles of Al and Al2O3:

375 g Al × (1 mol Al / 26.98 g Al) = 13.9 mol Al (3 significant figures)

According to the balanced equation, 4 moles of Al produce 2 moles of Al2O3. Therefore:

13.9 mol Al × (2 mol Al2O3 / 4 mol Al) = 6.95 mol Al2O3

Now, to find the mass of Al2O3 produced:

6.95 mol Al2O3 × (101.96 g Al2O3 / 1 mol Al2O3) = 708 grams of Al2O3 (3 significant figures)

Answer 2

To find the grams of Al2O3 produced, you need to calculate the moles of Al reacted, use the mole ratio from the balanced equation, and convert moles to grams using the molar mass of Al2O3. In this case, 375 grams of Al will produce 707 grams of Al2O3.

Step-by-step explanation:

To find the number of grams of Al2O3 produced, we first need to calculate the number of moles of Al reacted. Using the molar mass of Al (26.98 g/mol), we can convert the given mass of Al into moles:

375 g Al ÷ 26.98 g/mol Al = 13.9 mol Al

Then, we use the balanced chemical equation 4Al + 3O2 → 2Al2O3 to determine the mole ratio between Al and Al2O3. The ratio is 4:2, which means that for every 4 moles of Al reacted, 2 moles of Al2O3 are produced:

13.9 mol Al × (2 mol Al2O3 / 4 mol Al) = 6.95 mol Al2O3

Finally, we convert the moles of Al2O3 to grams using the molar mass of Al2O3 (101.96 g/mol):

6.95 mol Al2O3 × 101.96 g/mol Al2O3 = 707 g Al2O3

Therefore, 375 grams of Al will produce 707 grams of Al2O3.