How many grams of aluminum burned if 200.0 grams of aluminum oxide formed?

2 Answers
David Drayer
May 8, 2018

approximately 106 grams

Explanation:

The molecular mass of Aluminum Oxide is approximately 102 grams per mole.

# 2 xx Al = 2 xx 27 = 54 #
# 3 xx O = 3 xx 16 = 48 #
# 54 + 48 = 102 #

# 200/102 = 1.96 #moles

There are two atoms of Aluminum in one mole and each atom has a mass of approximately 27 grams so

Mass of Al = #1.96 xx 2 xx 27 = 106# grams

Meave60
May 9, 2018

#"105.9 g Al"# were burned in excess #"O"_2"# to produce #"200.0 g Al"_2"O"_3"#.

Explanation:

Balanced equation

#"4Al(s) + 3O"_2("g")stackrel(Delta")rarr##"2Al"_2"O"_3"#

There are three steps required to answer this question.

  1. Determine moles of #"Al"_2"O"_3"# by dividing the given mass by its molar mass #("101.960 g/mol")#. I prefer to do this by multiplying by the reciprocal of the molar mass (mol/g).

  2. Determine moles of #"Al"# by multiplying moles #"Al"_2"O"_3"# by the mole ratio between #"Al"# and #"Al"_2"O"_3"# from the balanced equation, with #"Al"# in the numerator.

  3. Determine mass of #"Al"# by multiplying moles #"Al"# by its molar mass #("26.982g/mol")#.

#200.0color(red)cancel(color(black)("g Al"_2"O"_3))xx(1"mol Al"_2"O"_3)/(101.960color(red)cancel(color(black)("g Al"_2"O"_3)))="1.962 mol Al"_2"O"_3"#

#1.962color(red)cancel(color(black)("mol Al"_2"O"_3))xx(4"mol Al")/(2color(red)cancel(color(black)("mol Al"_2"O"_3)))="3.924 mol Al"#

#3.924color(red)cancel(color(black)("mol Al"))xx(26.982"g Al")/(1color(red)cancel(color(black)("mol Al")))="105.9 g"# (rounded to four significant figures)

We can combine all steps into one equation:

#200.0color(red)cancel(color(black)("g Al"_2"O"_3))xx(1color(red)cancel(color(black)("mol Al"_2"O"_3)))/(101.960color(red)cancel(color(black)("g Al"_2"O"_3)))xx(4color(red)cancel(color(black)("mol Al")))/(2color(red)cancel(color(black)("mol Al"_2"O"_3)))xx(26.982"g Al")/(1color(red)cancel(color(black)("mol Al")))="105.9 g"# (rounded to four significant figures)

#"105.9 g Al"# were burned in excess #"O"_2"# to produce #"200.0 g Al"_2"O"_3"#.

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