# Question #2a7bc

May 12, 2016

$0.8$ moles

#### Explanation:

Given reaction and its balanced equation is

$2 {\text{Al"(s) + 3"FeO"(s) -> 3"Fe"(s) + "Al"_2"O}}_{3} \left(s\right)$

From the equation
$3$ moles of $\text{FeO}$ react completely with moles of $\text{Al} = 2$
$\therefore 1$ moles of $\text{FeO}$ reacts completely with moles of $\text{Al} = \frac{2}{3}$
or $1.2$ moles of $\text{FeO}$ react completely with moles of $\text{Al}$
$= \frac{2}{3} \times 1.2 = 0.8$

May 13, 2016

Looking at the mole ratio between FeO : Al, by looking at the equation, you can see that the mole ratio is 3 : 2.

So, Imagine if 3 moles of FeO gives you 2 moles of Al,

then 1.2 moles of Feo should give you:
[$1.2 \div 3 \times 2$] = 0.80 moles of Al.

Therefore, 0.80 moles of Al are needed to react completely with 1.2 moles of FeO.