JAMB - Chemistry (2006 - No. 35)

2C\(_2\)H\(_2\)(g) + 5O\(_2\)(g) → 4CO\(_2\)(g) + 2H\(_2\)O(g)
In the reaction above, the mass of carbon(IV)oxide produced on burning 78 g of ethyne is
[C =12, O = 16, H = 1]
264 g
39 g
352 g
156 g

Explanation

  2C\(_2\)H\(_2\)(g) + 5O\(_2\)(g) → 4CO\(_2\)(g) + 2H\(_2\)O(g)

  2 moles  :  4 moles

  n = \(\frac{mass}{molar mass}\) = \(\frac{78}{26}\)

 If  3 moles  :  6 moles 

 n = \(\frac{m}{M}\)

  m = n x M 

 m = 6 x 44

 m = 264g

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