Question #1f415

1 Answer
Nov 7, 2017

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