For reaction: 2KClO3(s) -----> 2KCl(s) + 3O2(g) PV=nRT or n= PV/RT P= .997 T=297

Question

For reaction: 2KClO3(s) -----> 2KCl(s) + 3O2(g)   PV=nRT or n= PV/RT

P= .997

T=297K

V=3.9L

R=.0821

n= mass(g) KClO3 Looking for mass of KClO3 that decomposed when 3.9L of O2 was collected when reaction took place at 23.8 degrees C and a P of 758torr.

I solve for n and get .160 mol of KClO3. Do I then multiply that times 2 (.32) before converting it to grams since there is a 2 in the balanced equation as the coeffient? Then continue to convert to grams (multiply x 122.55) for answer of 39.216 or 39.22g KClO3?

Wasn't sure what to do at the end here to get final mass.

Explanation / Answer

First, solve for the moles of O2 using PV = nRT:

n = PV / RT = (0.997 atm)(3.9 L) / (0.0821 atm*L/mol*K)(296.95 K)

n = 0.159 moles

Now, using the balanced chemical equation, solve for the moles KClO3:

0.159 of O2 x 2 moles of KClO3 / 3 moles of O2 = 0.106 moles

Now, solve fo the mass of KClO3 from its molar mass:

0.106 moles x 122.55 g/mol = 12.99 g

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