how many moles of o2 are needed to completely react with 54.0 g of al? 4al + 3o2 → 2al2o3 lets set up the…

how many moles of o2 are needed to completely react with 54.0 g of al? 4al + 3o2 → 2al2o3 lets set up the conversions first. what goes in the green box? 54.0 g al 3 mol o2 4 mol al 54.0 g al 26.98 g al

how many moles of o2 are needed to completely react with 54.0 g of al? 4al + 3o2 → 2al2o3 lets set up the conversions first. what goes in the green box? 54.0 g al 3 mol o2 4 mol al 54.0 g al 26.98 g al

Answer

Explanation:

Step1: Calculate moles of Al

The molar - mass of Al is approximately $26.98\ g/mol$. To find the number of moles of Al, we use the formula $n=\frac{m}{M}$, where $m = 54.0\ g$ and $M = 26.98\ g/mol$. So the first conversion factor is $\frac{1\ mol\ Al}{26.98\ g\ Al}$.

Step2: Use mole - ratio from the balanced equation

The balanced chemical equation is $4Al + 3O_2\rightarrow2Al_2O_3$. The mole - ratio of $Al$ to $O_2$ is $4:3$. So the second conversion factor is $\frac{3\ mol\ O_2}{4\ mol\ Al}$. In the green box, we need to convert grams of Al to moles of Al, so the value in the green box is $\frac{1\ mol\ Al}{26.98\ g\ Al}$.

Answer:

$\frac{1\ mol\ Al}{26.98\ g\ Al}$