Question

Question #1f415

Answer 1

Answer: 319 g of `Al_2O_3`

Explanation:

We know that at STP,

1 mol = 22.4 L

1. Convert 105 L of oxygen into number of moles:

22.4 L = 1 mol or,

105 L = `frac(1)(22.4)xx105` mol = 4.6875 mol

From the balanced chemical equation, we know that:

3 mol of `O_2` produces 2 mol of `Al_2O_3` or,

4.6875 mol of `O_2` produces `frac(2)(3)xx"4.6875 mol" = 3.125` mol of `Al_2O_3`.

2. Convert 3.125 mol of `Al_2O_3` into grams:

We know,

molar mass of `Al_2O_3` = 101.96 g/mol which means:

1 mole of `Al_2O_3` = 101.96 g or,

3.125 mol of `Al_2O_3 = 101.96xx3.125` g = 319 g of `Al_2O_3`