When a 12 g mixture og c and s burnt in air then a mix of co2 and so2 is produced in which the no of moles of so2 is half that of co2 the mass of c in the mix?
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C + O2 ---> CO2
S + O2 ---> SO2
n(C) = number of moles of carbon in the 12 gram mixture
n(S) = number of moles of sulphur in the 12 gram mixture
n(C)*12 + n(s)*32 = 12
n(C)/n(S) = 2
The first equation is the mass balance
The second equations is the C/S ratio = CO2/SO2 molar ratio
n(C) = n(S)*2
n(S)*24 + n(S)*32 = 12
n(S) = 0.214
n(C) = 0.214*2 = 0.428.
Mass of carbon in the mixture = 0.428*12 = 5.14 grams
S + O2 ---> SO2
n(C) = number of moles of carbon in the 12 gram mixture
n(S) = number of moles of sulphur in the 12 gram mixture
n(C)*12 + n(s)*32 = 12
n(C)/n(S) = 2
The first equation is the mass balance
The second equations is the C/S ratio = CO2/SO2 molar ratio
n(C) = n(S)*2
n(S)*24 + n(S)*32 = 12
n(S) = 0.214
n(C) = 0.214*2 = 0.428.
Mass of carbon in the mixture = 0.428*12 = 5.14 grams
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