# Calculate the mass of CO needed to react completely with 50.Og of Fe2 03?

Apr 6, 2018

$\approx 26.31$ grams

#### Explanation:

We must first create a balanced equation. Carbon monoxide is commonly used in industry to liberate iron from its oxides (which are often found in rock). This produced Carbon dioxide and solid iron.

$F {e}_{2} {O}_{3} \left(s\right) + 3 C O \left(g\right) \to 2 F e \left(s\right) + 3 C {O}_{2} \left(g\right)$

As $C O$ is used to liberate iron from Iron (III) Oxide.

Now we must find how many moles 50.0 g of Iron (III) oxide corresponds to:

$M$(iron (III) oxide)=$159.69 g m o {l}^{-} 1$
$m = 50.0$
$n$=?

Using the equation:

$n = \frac{m}{M}$

$n = \frac{50.0}{159.69}$

$n \approx 0.3131$

Now from the molar ratios we can see that we require 3 moles of $C O$ for every 1 mole of Iron (III) oxide.

$n \left(C O\right) = 3 \times 0.3131 = 0.9393$

Then using the same equation we can find the mass of $C O$ required:

$n = \frac{m}{M}$
$M \left(C O\right) = 28.01 g m o {l}^{-} 1$

$0.9393 = \frac{m}{28.01}$

$m \approx 26.31 g$ of $C O$ is needed.