# Question f3642

Nov 8, 2015

The percent composition of nitrogen in ammonium nitrate is $\text{34.99% N}$.

#### 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 $\text{80.052 g/mol}$.

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%"

Nov 8, 2015

$\text{1.00 kg Fe"_2"O"_3}$ will produce $\text{827 g CO"_2}$ in this reaction.

#### Explanation:

Balanced Equation

$\text{Fe"_2"O"_3("s")" + 3CO(g)}$$\rightarrow$$\text{2Fe(s)" + "3CO"_2("g")}$

On a previous question with the same equation, you indicated that the amount of hematite $\left(\text{Fe"_2"O"_3}\right)$ is $\text{1.00 kg}$.

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

$\text{Fe"_2"O"_3} :$(2xx55.845 "g/mol")+(3xx15.999 "g/mol")="159.687 g/mol Fe"_2"O"_3"

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

DETERMINE THE MASS OF $\text{CO"_2}$ IN GRAMS

Convert $\text{1.00 kg Fe"_2"O"_3}$ to grams.

$1.00 \cancel{\text{kg"xx(1000"g")/(1cancel"kg")="1000 g Fe"_2"O"_3}}$

Determine the moles of $\text{Fe"_2"O"_3}$ by dividing its mass by its molar mass.

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

Determine the moles $\text{CO"_2}$ that can be produced by multiplying moles $\text{Fe"_2"O"_3}$ times the mole ratio between $\text{Fe"_2"O"_3}$ and $\text{CO"_2}$.

$6.2623 \cancel{\text{mol Fe"_2"O"_3xx(3"mol CO"_2)/(1cancel"mol Fe"_2"O"_3)="18.7869 mol CO"_2}}$

Determine the mass of $\text{CO"_2}$ produced by multiplying the moles $\text{CO"_2}$ times its molar mass.

$18.7869 \cancel{\text{mol CO"_2xx(44.001"g CO"_2)/(1cancel"mol CO"_2)="827 g CO"_2}}$

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