# Question e3c64

Jul 6, 2017

$17.5$ $\text{kg Fe}$

#### Explanation:

We're asked to find the theoretical yield, in $\text{kg}$, of $\text{Fe} \left(s\right)$ that can be produced by reacting $25.0$ ${\text{kg Fe"_2"O}}_{3} \left(s\right)$.

We can solve this problem by first converting the given mass of ${\text{Fe"_2"O}}_{3}$ to moles using its molar mass, calculated to be $159.69$ $\text{g/mol}$. (We need to convert from kilograms to grams as well):

25.0cancel("kg Fe"_2"O"_3)((10^3cancel("g"))/(1cancel"kg"))((1color(white)(l)"mol Fe"_2"O"_3)/(159.69cancel("g Fe"_2"O"_3))) = color(red)(156.56 color(red)("mol Fe"_2"O"_3

Using the coefficients of the chemical equation, we can find the relative number of moles of $\text{Fe}$ produced:

color(red)(156.56)cancel(color(red)("mol Fe"_2"O"_3))((2color(white)(l)"mol Fe")/(1cancel("mol Fe"_2"O"_3))) = color(green)(313.11 color(green)("mol Fe"

Lastly, we can use the molar mass of iron ($55.85$ $\text{g/mol}$) to find the number of grams, and then kilograms, of $\text{Fe}$ that can be produced:

color(green)(313.11)cancel(color(green)("mol Fe"))((55.85cancel("g Fe"))/(1cancel("mol Fe")))((1color(white)(l)"kg")/(10^3cancel("g"))) = color(blue)(17.5 color(blue)("kg Fe"#

rounded to $3$ significant figures.