Answer:
Explanation:
in a combustion of ethane 2 moles of ethane react with 7 moles of O2
now no of moles in 54 gram of O2=mass/ molar mass
moles =54/32=1.7 moles
if 7 moles of O2 required 2 moles of ethane then 1.7 mole required=?
7 moles of O2=2 moles of C2H6
1.7 moles of O2=1.7*2/7=0.5 moles of C2H6
0.5 moles of C2H6 contain how much grams=?
mass= moles*molar mass=0.5*30=15
