Question

How much potassium chlorate is needed to produce 20.0 mL ofoxygen gas at 670 mm Hg and 293 K?

Answer 1

Answer:

0.058824 g

Explanation:

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

$PV=nRT$

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 = $62.3637\text{ L.mmHg }mol^{-1}K^{-1}$

n = number of moles of oxygen gas = ?

Putting values in above equation, we get:

$670mmHg\times 0.02L=n\times 62.3637\text{ L.mmHg }mol^{-1}K^{-1}\times 293K\\\\n=\frac{670\times 0.02}{62.3637\times 293}=0.00073mol$

According the reaction shown below:-

$2KClO_3\rightarrow 2KCl+3O_2$

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

So,

1 mole of oxygen gas is produced when $\frac{2}{3}$ moles of potassium chlorate reacts

Also,

0.00073 mole of oxygen gas is produced when $\frac{2}{3}\times 0.00073$ 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 = $0.00048\times 122.55\ g$ = 0.058824 g