# A 5*g mass of aluminum metal is oxidized by a 10*g mass of sulfur. If a 12.5*g mass of "aluminum sulfide" is isolated, what is the percentage yield?

Dec 30, 2017

Aluminum is the limiting reagent....

#### Explanation:

We assess the following redox reaction....

$2 A l \left(s\right) + 3 S \left(s\right) \stackrel{\Delta}{\rightarrow} A {l}_{2} {S}_{3}$

$\text{Moles of aluminum metal} = \frac{5.00 \cdot g}{27.0 \cdot g \cdot m o {l}^{-} 1} = 0.185 \cdot m o l$

$\text{Moles of sulfur} = \frac{10.00 \cdot g}{32.06 \cdot g \cdot m o {l}^{-} 1} = 0.312 \cdot m o l$

And thus the sulfur is in stoichiometric excess, and at most we can make $\frac{0.185 \cdot m o l}{2}$ with respect to $\text{aluminum sulfide}$ $\equiv \frac{0.185 \cdot m o l}{2} \times 150.16 \cdot g \cdot m o {l}^{-} 1 = 13.9 \cdot g$..

And thus "% yield"=(12.5*g)/(13.9*g)xx100%=??%

I leave it to you to calculate the mass of excess sulfur.