Question

Question #55f2f

Answer 1

`5.37` `"g CO"_2`

Explanation:

We're asked to find the mass (in grams) of gas that will form if `13.00` `"g Fe"_2"O"_3` and `4.20` `"g C"` react.

Let's first write the balanced chemical equation for this reaction:

`ul(2"Fe"_2"O"_3 (s) +3 "C"(s) rarr 4"Fe"(s) + 3"CO"_2(g)`

Carbon dioxide is the gas in question, so we will be finding the number of grams of `"CO"_2`.

What we need to do here is find the limiting reactant of this reaction. To do so, we follow these steps:

1. Find the number of moles of each reactant present (using molar mass)

2. Divide the mole values by the coefficient in the equation

3. Whichever value is lowest is the limiting reactant

1.

Let's find the number of moles of each reactant, using the molar masses of each:

`13.00cancel("g Fe"_2"O"_3)((1color(white)(l)"mol Fe"_2"O"_3)/(159.70cancel("g Fe"_2"O"_3))) = ul(0.0814color(white)(l)"mol Fe"_2"O"_3`

`4.20cancel("g C")((1color(white)(l)"mol C")/(12.011cancel("g C"))) = ul(0.350color(white)(l)"mol C"`

2.

Now, we divide each value by the respective coefficient in the chemical equation

`(0.0814color(white)(l)"mol Fe"_2"O"_3)/(2color(white)(l)"(coefficient)") = ul(0.0407`

`(0.350color(white)(l)"mol C")/(3color(white)(l)"(coefficient)") = ul(0.117`

3.

Since the value for iron(III) oxide is lower, `"Fe"_2"O"_3` is the limiting reactant.

What we do now is use the mole value of `"Fe"_2"O"_3` (`0.0814color(white)(l)"mol"`) and the coefficients of the equation to calculate the relative number of moles of `"CO"_2` that can form:

`0.0814cancel("mol Fe"_2"O"_3)((3color(white)(l)"mol CO"_2)/(2cancel("mol Fe"_2"O"_3))) = color(red)(ul(0.122color(white)(l)"mol CO"_2`

Lastly, we use the molar mass of carbon dioxide (`44.009` `"g/mol"`) to calculate the mass (in grams) that can form:

`color(red)(0.122)cancel(color(red)("mol CO"_2))((44.009color(white)(l)"g CO"_2)/(1cancel("mol CO"_2))) = color(blue)(ulbar(|stackrel(" ")(" "5.37color(white)(l)"g CO"_2" ")|)`