Given the equation 2KCIO3(s)=2KCI(s) + 3O2(g). A 3.00-g sample of KCIO3 is decomposed and the oxygen at 24 degrees C and 0.982 atm is collected. What volume of oxygen gas will be collected assuming 100% yield?

0.912 L or 912 mL

Explanation:

 M(KClO3) =  122.55 g/mol

3.00 g KClO3 * 1  mol/122.55 g = 3.00/122.55 mol =0.02449 mol                

                           2KCIO3(s)=2KCI(s) + 3O2(g)

from reaction      2 mol                         3 mol

given                   0.02449 mol              x

x = 0.02449*3/2 =0.03673 mol O2

T = 24 + 273.15 = 297.15 K

PV = nRT

V= nRT/P = (0.03673 mol*0.082057 L*atm/K*mol*297.15 K)/0.982 atm =

= 0.912 L or 912 mL

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