Question

Solid aluminum (AI) and oxygen (0) gas react to form solid aluminum oxide (AIO). Suppose you have 7.0 mol of Al and 13.0 mol of o, in a reactor. Suppose as much as possible of the Al reacts. How much will be left? Round your answer to the nearest 0.1 mol mol 0.

Answer 1

Answer:

$n_{O_2}^{leftover}=7.7mol$

Explanation:

Hello there!

In this case, according to the given information, it turns out firstly necessary for us to set up the corresponding chemical equation:

$4Al+3O_2\rightarrow 2Al_2O_3$

In such a way, we calculate the moles of aluminum consumed by 13.0 moles of oxygen in the reaction, by applying the 4:3 mole ratio between them:

$n_{Al}=13.0molO_2*\frac{4molAl}{3molO_2} =17.3molAl$

This means that Al is actually the limiting reactant and oxygen is in excess, for that reason we calculate the moles of oxygen consumed by 7.0 moles of aluminum:

$n_{O_2}=7.0molAl*\frac{3molO_2}{4molAl} =5.3molO_2$

Thus, the leftover of oxygen is:

$n_{O_2}^{leftover}=13.0mol-5.3mol\\\\n_{O_2}^{leftover}=7.7mol$

Whereas all the aluminum is assumed to be consumed.

Regards!