# Question #6cb30

Mar 10, 2015

You'd need $\text{20.9 g}$ of iron to react with that much oxygen.

Start by taking a look at your balanced chemical equation

$3 F e + 2 {O}_{2} \to F {e}_{3} {O}_{4}$

Notice that you have a $\text{3:2}$ mole ratio between iron and oxygen - this represents the ratio according to which iron and oxygen will mix to form the product, $F {e}_{3} {O}_{4}$ (commonly known as magnetite).

So, for every 3 moles of iron, you need 2 moles of oxygen for the reaction to take place. You can determine the number of moles of oxygen by using its molar mass

$\text{8.0 g oxygen" * "1 mole"/"32.0 g" = "0.250 moles oxygen}$

This much oxygen will need

$\text{0.250 moles oxygen" * "3 moles iron"/"2 moles oxygen" = "0.375 moles iron}$

Using iron's molar mass to determine how many grams are needed will produce

$\text{0.375 moles iron" * "55.85 g"/"1 mole" = "20.9 g iron}$