Question

if equal weight of p and o2 were heated to form p203 and p205 in molar ratio 2:1 ​

Answer 1

4 P + 3 O2 → 2 P2O3  

4 P + 5 O2 → 2 P2O5  

Since you are told that the result is equal moles of P2O3 and P2O5, these two equations can be added together to form the overall reaction:  

8 P + 8 O2 → 2 P2O3 + 2 P2O5  

Then simplifying:  

4 P + 4 O2 → P2O3 + P2O5  

Take hypothetical samples of 100 grams each of P and O2:  

(100 g P) / (30.97376 g P/mol) = 3.2285 mol P  

(100 g O2) / (31.99886 g O2/mol) = 3.1251 mol O2  

3.1251 moles of O2 would react completely with 3.1251 x (4/4) = 3.1251 moles of P, but there is more P present than that, so P is in excess and O2 is the limiting reactant (component), meaning that O2 is exhausted and P is left over.  

(3.2285 mol P initially) - (3.1251 mol P reacted) = 0.1034 mol P left over  

(0.1034 mol P) / (3.2285 mol P initially) = 0.0320 = 3.20% P left over

Hope this helps!

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Answer 2

Answer:

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