Question

what is the n-factor of C3H8 in its combustion with oxygen?​

Answer 1

The nth factor of C3H8 is 20 , for Co2 is 20/3 , for O2 is 4 and for H20  is 5

Explanation:

When C3H8 is combined with oxygen we get the reaction is

$C_{3} H_{8} + O_{2}$  -> $Co_{2} + H_{2} O$

Balancing the equation

$C_{3} H_{8} + 5 O_{2} - > 3C o_{2} + 4H_{2}O$

$6H_{2}O + C_{3} H_{8} - > 3Co_{2}+20 H^{+} +20 e^{-}$

$[4H^{+} + 4e^{-} + O_{2} - > H_{2} O+H_{2}O] 5$

Adding two equation, we get the final equation.

For 1 mole of $C_{3} H_{8} - > 20e^{-}$ are involved

$n_{f}$  for $C_{3}H_{8} = 20$

For 3 mole of $Co_{2} - > 20e^{-}$   are involved

For 1 mole of $Co_{2} - > \frac{20}{3} e^{-}$  are involved

$n_{f}$  for  $Co_{2} = \frac{20}{3}$

$n_{f}$  for  $O_{2} = \frac{20}{5} = 4$

$n_{f}$  for  $H_{2}O = \frac{20}{4} = 5$

Final answer:

The nth factor of C3H8 is 20 , for carbon dioxide (Co2) is 20/3 , for oxygen (O2) is 4 and for water (H20) is 5

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