# What is the maximum number of mols of copper (III) sulfide that can be formed when 8.0 mols of copper reacts with 9.0 mols of sulfur?

Mar 20, 2016

You have to find the reaction, balance it and then compare the mole ratios.

$5 , \overline{33} g$

#### Explanation:

The chemical reaction is:

$4 C u + 3 {S}_{2} \to 2 C {u}_{2} {S}_{3}$

So from the left part of the equation, once can say:

For every 4 moles of copper, 3 moles of sulfur is needed.
For every 8 moles of copper, $x$ moles of sulfur are needed.
For every $y$ moles of copper, 9 moles of sulfur are needed.

$x = \frac{3 \cdot 8}{4} = 6$ moles of sulfur We have 9, which is enough

$y = \frac{4 \cdot 9}{3} = 12$ moles of copper We have 8, which is not enough

So finally, 8 moles of sulfur will react. The reaction will become:

$4 C u + 3 {S}_{2} \to 2 C {u}_{2} {S}_{3}$

$z C u + 8 {S}_{2} \to w C {u}_{2} {S}_{3}$

$z = 4 \cdot \frac{8}{3} = \frac{32}{3}$

$w = 2 \cdot \frac{8}{3} = \frac{16}{3}$

$\frac{32}{3} C u + 8 {S}_{2} \to \frac{16}{3} C {u}_{2} {S}_{3}$

So the maximum number moles that can be formed are:

$\frac{16}{3} = 5 , \overline{33} g$