How many grams of Al2O3 are needed to produce 24.5 L of O2 gas at STP?

74.35 g

Explanation:

The equation  of the reaction is given as;

2Al2O3 --> 4Al + 3O2

From the reaction;

2 mol of Al2O3 reacts to produce 3 mol of O2

Converting 2 mol of Al2O3 to mass;

Number of moles = Mass / Molar mass

Mass = Number of moles * Molar mass

Mass = 2 * 101.96 = 203.92 g

Converting 3 mol of O2 to volume;

1 mol = 22.4 L

3 mol = x

x = 22.4 * 3 = 67.2 L

This means 203.92g of Al2O3 produces 67.2L of O2

xg would produce 24.5 L of O2

203.92 = 67.2

x = 24.5

x = (24.5 * 203.92) / 67.2

x = 74.35 g

pie chart because it is in percents ( % )