Answer to Question #171592 in Inorganic Chemistry for Halima Abudu

Question #171592

2g of a mixture of KCLO3 and KCL yielded 336cm^3 of O2 measured dry at 747mmHg on heating. Calculate the volume of dry O2 measured at stp and the percentage composition of KCLO3.


1
Expert's answer
2021-03-17T06:44:59-0400

2KClO3 → 2KCl + 3O2

We will use the temperature value of 127 ºC.

T = 127 + 273 = 400 K

P = 747 mmHg = 0.982 atm

V = 336 mL = 0.336 L

Ideal Gas Law

PV = nRT

R = 0.08206 L×atm/mol×K

At STP one mole of any gas occupies volume of 22.4 L.

At STP

According to the reaction:

n(KClO3)

M(KClO3) = 122.55 g/mol

Proportion:

2 g – 100 %

0.817 g – x



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