What mass of potassium chlorate should be required to produce 113 L of oxygen O2. measured at STP g

411.6 g

Explanation:

The equation for the decomposition of potassium chlorate to produce oxygen is given as;

2KClO3 --> 2KCl + 3O2

From the stochiometry of the reaction 2 mole of KClO3 produces 3 mol of O2

At STP,

1 mol = 22.4L

3 mol = x

x = 67.2L

2 mole of KClO3 produces 67.2L of O2

How many moles would produce 113L?

2 = 67.2

x = 113

x = 3.36 moles

Moles can be converted to mass using;

Mass = moles * Molar mass

Mass = 3,36 * 122.5 g/mol

Mass = 411.6 g

the answer is definitely c.