Question

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Answer 1

Answer:

d. 12.3 grams of Al2O3

Explanation:

The balanced equation in this question is as follows:

4Al + 3O2 → 2Al2O3

In this equation above, 4 moles of Al produces 2 moles of Al2O3

However, the moles of Al2O3 must first be found using;

mole = mass/molar mass

Molar mass of Al = 27g/mol

mole = 6.50/27

= 0.241mol of Al.

Hence, if 4 moles of aluminum (limiting reagent) reacts to form 2 moles of aluminum oxide (Al2O3).

Then, 0.241mol of Al will produce (0.241 × 2/4) = 0.241/2 = 0.121mol of Al2O3.

Convert this mole value to molar mass using mole = mass/molar mass

Molar mass of Al2O3 = 27(2) + 16(3)

= 54 + 48

= 102g/mol

mass = molar mass × mole

mass = 102 × 0.121

mass of Al2O3 = 12.34grams.