Gaseous ethane CH3CH3 will react with gaseous oxygen O2 to produce gaseous carbon dioxide CO2 and gaseous water H2O . Suppose 0.30 g of ethane is mixed with 2.20 g of oxygen. Calculate the maximum mass of carbon dioxide that could be produced by the chemical reaction. Round your answer to 2 significant digits.
equetion; 2CH3CH3(g) + 7O2(g) -----> 4CO2(g) + 6 H2O(g)
0.3 g 2.20 g (?)g
Ethane mol. mass=30 g
O2 Mol. mass= 32 g
CO2 mol. mass= 44 g
0.3 g Ethane is 0.3/30 = 0.01 mole ; 2.20 g O2 is 2.20/32= 0.0687 mole
from equetion,1 mole Ethane required 3.5 mole O2
so, 0.01 mole Ethane required= 0.01*3.5=0.035 mole O2
we have enough amount of O2, so, Ethane is limiting reagent
1 mole Ethane produce 2 mole CO2
0.01 mole Ethane produce = 0.01*2= 0.02 mole CO2 =0.88 g CO2
given reaction produce 0.88 g Carbon dioxide
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