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A 6.50 gram piece of aluminum reacts with an excess of oxygen. use the balanced equation below to determine how many grams of aluminum oxide is formed during this reaction.
4 Al + 3 O2 –> 2 Al2O3
a. 662.7 grams of Al2O3
b. 6.1 grams of Al2O3
c. 24.6 grams of Al2O3
d. 12..3 grams of Al2O3
Question
Answer:
d. 12.3 grams of Al2O3
Explanation:
The balanced chemical equation of this chemical reaction is as follows:
4Al + 3O2 –> 2Al2O3
Based on the balanced equation, 4 moles of aluminum (limiting reagent) reacts to form 2 moles of aluminum oxide (Al2O3).
First, we need to convert the mass of aluminum to moles using the formula;
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.