Question

Question #f3642

Answer 1

The percent composition of nitrogen in ammonium nitrate is `"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"`.

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%"`

Answer 2

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

Explanation:

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"`)