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Mole Concept in Balanced Reaction

What weights ...

Question

What weights of P₄O₆and P₄O₁₀ will be produced by the combustion of 2gof P₄in 2gof oxygen leaving no P₄ and O₂ ?

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Solution

$E q u a t i o n s t o b e c o n s i d e r e d a s : P_{4} + 3 O_{2} \to P_{4} O_{6} P_{4} + 5 O_{2} \to P_{4} O_{10} L e t x b e t h e m a s s o f P_{4} t h a t i s c o n v e r t e d i n t o P_{4} O_{6} a n d t h e r e s t (2 - x) g i s c o n v e r t e d t o P_{4} O_{10} . M a s s o f o x y g e n r e q u i r e d f o r P_{4} O_{6} = \frac{x}{31 \times 4} \times 96 = \frac{96}{124} x g M a s s o f o x y g e n r e q u i r e d o f P_{4} O_{10} = \frac{(2 - x)}{31 \times 4} \times 160 = \frac{160}{124} (2 - x) g T h e t o t a l m a s s o f o x y g e n u s e d = 2 g S o, \frac{96}{124} x + \frac{160}{124} (2 - x) = 2 96 x + 320 - 160 x = 248 320 - 248 = (160 - 96) x x = \frac{72}{64} = 1.125 g M a s s o f P_{4} O_{6} o b t a i n e d = \frac{M a s s o f P_{4}}{M o l a r m a s s o f P_{4}} \times m o l a r m a s s o f P_{4} O_{6} = \frac{1.125}{124} \times 220 = 1.996 g M a s s o f P_{4} O_{10} o b t a i n e d = T o t a l m a s s t a k e n - M a s s o f P_{4} O_{6} = 4 - 1.996 = 2.004 g$

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