Question #f3642

To find the percent composition of a particular element in a compound, you first must find the molar mass of the compound you have. In this case the molar mass of ammonium nitrate is #"80.052 g/mol"#.

You are looking for the percent composition of N. You have two moles of N in the compound, so you must multiply the molar mass of N by 2. #(2xx14.007"g/mol")="28.014 g/mol"#

You then divide the mass of N by the mass of the molecule:
#(28.014 "g N")/(80.052 "g NH"_4"NO"_3)xx100="34.99%"#

Balanced Equation

#"Fe"_2"O"_3("s")" + 3CO(g)"##rarr##"2Fe(s)" + "3CO"_2("g")"#

On a previous question with the same equation, you indicated that the amount of hematite #("Fe"_2"O"_3")# is #"1.00 kg"#.

Determine the Molar Masses of #"Fe"_2"O"_3"# and #"CO"_2"# Multiply the subscripts times the molar mass of each element (atomic weight on the periodic table in g/mol), and add them.

#"Fe"_2"O"_3":##(2xx55.845 "g/mol")+(3xx15.999 "g/mol")="159.687 g/mol Fe"_2"O"_3"#

#"CO"_2":##(1xx12.0107"g/mol")+(2xx15.999"g/mol")="44.001 g/mol CO"_2"#

DETERMINE THE MASS OF #"CO"_2"# IN GRAMS

Convert #"1.00 kg Fe"_2"O"_3"# to grams.

#1.00cancel"kg"xx(1000"g")/(1cancel"kg")="1000 g Fe"_2"O"_3"#

Determine the moles of #"Fe"_2"O"_3"# by dividing its mass by its molar mass.

#1000cancel"g Fe"_2"O"_3xx(1"mol Fe"_2"O"_3)/(159.687cancel"g Fe"_2"O"_3)="6.2623 mol Fe"_2"O"_3"#
I am keeping some guard units to reduce rounding errors. I will round to three significant figures at the end.

Determine the moles #"CO"_2"# that can be produced by multiplying moles #"Fe"_2"O"_3"# times the mole ratio between #"Fe"_2"O"_3"# and #"CO"_2"#.

#6.2623 cancel"mol Fe"_2"O"_3xx(3"mol CO"_2)/(1cancel"mol Fe"_2"O"_3)="18.7869 mol CO"_2"#

Determine the mass of #"CO"_2"# produced by multiplying the moles #"CO"_2"# times its molar mass.

#18.7869 cancel"mol CO"_2xx(44.001"g CO"_2)/(1cancel"mol CO"_2)="827 g CO"_2"#

These steps can be combined into one step.
#1000cancel"g Fe"_2"O"_3xx(1cancel"mol Fe"_2"O"_3)/(159.687cancel"g Fe"_2"O"_3)xx(3cancel"mol CO"_2)/(1cancel"mol Fe"_2"O"_3)xx(44.001"g CO"_2)/(1cancel"mol CO"_2)="827 g CO"_2"# (rounded to three significant figures due to #"1.00 kg Fe"_2"O"_3"#)