Question #f3642

2 Answers
Nov 8, 2015

The percent composition of nitrogen in ammonium nitrate is "34.99% N"34.99% N.

Explanation:

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"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"(2×14.007g/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%"28.014g N80.052g NH4NO3×100=34.99%

Nov 8, 2015

"1.00 kg Fe"_2"O"_3"1.00 kg Fe2O3 will produce "827 g CO"_2"827 g CO2 in this reaction.

Explanation:

Balanced Equation

"Fe"_2"O"_3("s")" + 3CO(g)"Fe2O3(s) + 3CO(g)rarr"2Fe(s)" + "3CO"_2("g")"2Fe(s)+3CO2(g)

On a previous question with the same equation, you indicated that the amount of hematite ("Fe"_2"O"_3")(Fe2O3) is "1.00 kg"1.00 kg.

Determine the Molar Masses of "Fe"_2"O"_3"Fe2O3 and "CO"_2"CO2 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":Fe2O3:(2xx55.845 "g/mol")+(3xx15.999 "g/mol")="159.687 g/mol Fe"_2"O"_3"(2×55.845g/mol)+(3×15.999g/mol)=159.687 g/mol Fe2O3

"CO"_2":CO2:(1xx12.0107"g/mol")+(2xx15.999"g/mol")="44.001 g/mol CO"_2"(1×12.0107g/mol)+(2×15.999g/mol)=44.001 g/mol CO2

DETERMINE THE MASS OF "CO"_2"CO2 IN GRAMS

Convert "1.00 kg Fe"_2"O"_3"1.00 kg Fe2O3 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")