First you want to create and balance the chemical equation.

___KClO3--->___KCL+___O2

Since the K's and Cl's are already balanced, balance the O's.

2_KClO3--->___KCL+3__O2

Now balance the K's and Cl's

2__KClO3--->2_KCl+3_O2

The numbers you just added are the number of moles that take place in each equation. So for every 2 moles of KClO3 decomposed, 3 moles of O2 are created. This isn't important for the next step but remember it for the last.

The next step is to figure out how many moles of O2 are being produced. For this, you use the equation pv=nrt.

P(ressure)=0.832 atm

V(olume)=178L

n (moles)=Unknown

R = 0.08206 L·atm/mol·K

T(emperature)=431K (°C+273)

Substituting the values, you get

0.832 atm*178L=n*0.08206 L·atm/mol·K*431K

Which becomes

148.096=35.3851n

You then divide each side by 35.3851 to get n alone.

n=4.185

So 4.185 moles of O2 would be produced.

Which brings us back to the chemical equation. For every 2 moles of KClO3 decomposed, 3 moles of O2 are created.

Using dimensional reasoning, you can convert the calculated moles of O2 to moles of KClO3 needed.

4.185 moles O2*2moles KClO3/3 moles O2.

The moles of O2 cancel out, leaving you with moles of KClO3.

4.185*2=8.37

8.37/3=2.79

So you need 2.79 moles of KClO3

From here, you get the mass of one mole of KClO3, which is 122.548g.

From here, you use dimensional reasoning again.

2.79 moles*122.548g/mol=341.909g KClO3.

341.909g KClO3 is needed.

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PV = nRT

n = PV/RT

put in your P, V, R and T results to find n, the number of moles of O2.

KClO3 ---> KCl + O2

so moles of O2 produced = moles KClO3 decomposed.

From RFM of KClO3 convert this number of moles to grams.