Answer: Given for oxygen gas volume V=626 cm3=626 mL=0.626 L Pressure P=0.880 at

Question

Answer:

Given for oxygen gas volume V=626 cm3=626 mL=0.626 L

Pressure P=0.880 atm and temperature T=133°C=133+273 K

T=406 K.

From Ideal gas equation PV=nRT

Where R=gas constant=0.0821 L atm mol-1 K-1

n=PV/RT=(0.88 atm x 0.626 L)/(0.0821 L atm mol-1 K-1 x 406 K)

n=0.0165 mol.

Given reaction is 2KClO3(s)------->2 KCl(s)+3O2(g)

From the equation the mole ratio between KClO3 and O2 is 2:3, then moles of KClO3=2/3(mol O2)=2/3(0.0165 mol)

Mol KClO3=0.011

Molar mass of KClO3=122.55 g/mol

Mass of KClO3=molesxmolar mass=0.011 molx122.55 g/mol=1.35 g.

The mass of KClO3=1.35 g.

Explanation / Answer

What mass of KCIOs must be decomposed to produce 626 cm3 of oxygen gas at 133°C and 0.880 atm? (The other reaction product is solid KCL.) 2 KCIO:(s) 4. 2 KCI(s) +3 02(g)

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