Question #f3642

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Question #f3642

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Answer 1 · 2015-11-08T18:11:15

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. $(2 \times 14.007 g/mol) = 28.014 g/mol$ $(2xx14.007"g/mol")="28.014 g/mol"$

You then divide the mass of N by the mass of the molecule:
$\frac{28.014 g N}{80.052 {g NH}_{4} {NO}_{3}} \times 100 = 34.99%$ $(28.014 "g N")/(80.052 "g NH"_4"NO"_3)xx100="34.99%"$

Answer 2 · 2015-11-08T20:13:54

${1.00 kg Fe}_{2} O_{3}$ $"1.00 kg Fe"_2"O"_3"$ will produce ${827 g CO}_{2}$ $"827 g CO"_2"$ in this reaction.

Explanation:

Balanced Equation

${Fe}_{2} O_{3} (s) + 3CO(g)$ $"Fe"_2"O"_3("s")" + 3CO(g)"$ $\to$ $rarr$ $2Fe(s) + {3CO}_{2} (g)$ $"2Fe(s)" + "3CO"_2("g")"$

On a previous question with the same equation, you indicated that the amount of hematite $({Fe}_{2} O_{3})$ $("Fe"_2"O"_3")$ is $1.00 kg$ $"1.00 kg"$ .

Determine the Molar Masses of ${Fe}_{2} O_{3}$ $"Fe"_2"O"_3"$ and ${CO}_{2}$ $"CO"_2"$ Multiply the subscripts times the molar mass of each element (atomic weight on the periodic table in g/mol), and add them.

${Fe}_{2} O_{3} :$ $"Fe"_2"O"_3":$ $(2 \times 55.845 g/mol) + (3 \times 15.999 g/mol) = {159.687 g/mol Fe}_{2} O_{3}$ $(2xx55.845 "g/mol")+(3xx15.999 "g/mol")="159.687 g/mol Fe"_2"O"_3"$

${CO}_{2} :$ $"CO"_2":$ $(1 \times 12.0107 g/mol) + (2 \times 15.999 g/mol) = {44.001 g/mol CO}_{2}$ $(1xx12.0107"g/mol")+(2xx15.999"g/mol")="44.001 g/mol CO"_2"$

DETERMINE THE MASS OF ${CO}_{2}$ $"CO"_2"$ IN GRAMS

Convert ${1.00 kg Fe}_{2} O_{3}$ $"1.00 kg Fe"_2"O"_3"$ 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 $"1.00 kg Fe"_2"O"_3"$ )

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Question #f3642

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