Molar mass of C4H10,
MM = 4*MM(C) + 10*MM(H)
= 4*12.01 + 10*1.008
= 58.12 g/mol
mass(C4H10)= 2.3 g
use:
number of mol of C4H10,
n = mass of C4H10/molar mass of C4H10
=(2.3 g)/(58.12 g/mol)
= 3.957*10^-2 mol
Molar mass of O2 = 32 g/mol
mass(O2)= 11.6 g
use:
number of mol of O2,
n = mass of O2/molar mass of O2
=(11.6 g)/(32 g/mol)
= 0.3625 mol
Balanced chemical equation is:
2 C4H10 + 13 O2 ---> 8 CO2 + 10 H2O
2 mol of C4H10 reacts with 13 mol of O2
for 3.957*10^-2 mol of C4H10, 0.2572 mol of O2 is required
But we have 0.3625 mol of O2
so, C4H10 is limiting reagent
Since CH10 is limiting reagent, there should be no butane left over.
Answer: 0.0 g
(сн, сH), сн-) (CH will react with gaseous oxygen (0, to produce gaseous carbon dioxide CO2...
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