Question

Using the equation below, calculate the approximate moles of oxygen gas (O2) required to completely react with 719.68 moles of aluminum (Al). 4Al (s) + 3O2(g) -----> 2Al2O3(s)

Answer 1

Answer:

$n_{O_2}=959.57molO_2$

Explanation:

Hello,

In this case, given the reaction:

$4Al (s) + 3O_2(g) \rightarrow 2Al_2O_3(s)$

Since aluminum and oxygen are in a 4:3 molar ratio, we compute the moles of oxygen that completely react as shown below:

$n_{O_2}=719.68molAl*\frac{3molO_2}{4molAl} \\\\n_{O_2}=959.57molO_2$

Best regards.

Answer 2

Answer:

539.76 moles of O₂

Explanation:

Equation of reaction:

4Al + 3O₂ → 2Al₂O₃

From the equation of reaction above, 4 moles of Al will react with 3 moles of O₂

We can use mole-concept to find the number of moles of O₂.

4 moles of Al = 3moles of O₂

719.68 moles of Al = x moles of O₂

X = (3 × 719.68) / 4

X = 2159.04 / 4

X = 539.76 moles

539.76 moles of O₂ will react with 719.68 moles of Al