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]
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
There is an explanation video available below.
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