How much potassium chlorate is needed to produce 20.0 ml
ofoxygen gas at 670 mm hg and 293 k?

0.058824 g

Explanation:

To calculate the moles of oxygen gas formed, we use the equation given by ideal gas which follows:

where,

P = pressure of the gas = 670 mmHg  

V = Volume of the gas = 20.0 mL = 0.02 L ( 1 mL= 0.001 L)

T = Temperature of the gas =293 K

R = Gas constant =

n = number of moles of oxygen gas = ?

Putting values in above equation, we get:

According the reaction shown below:-

3 moles of oxygen gas is produced when 2 moles of potassium chlorate reacts

So,

1 mole of oxygen gas is produced when moles of potassium chlorate reacts

Also,

0.00073 mole of oxygen gas is produced when moles of potassium chlorate reacts

Moles of potassium chlorate reacts = 0.00048 Moles

Molar mass of potassium chlorate = 122.55 g/mol

Mass = Moles*Molar mass = = 0.058824 g