If 3.50 mol of ethane, CzH6, undergo combustion according to the unbalaced equation given of oxygen requires ? How many moles of each product formed?

The Balance Chemical equation for said reaction is as follow,

                        C₂H₆   + 3.5 O₂    →    2 CO₂  +  3 H₂O

Or,

                       2 C₂H₆   + 7 O₂    →    4 CO₂  +  6 H₂O

According to equation,

       2 moles of Ethane react completely with  =  7 moles of Oxygen

So,

              4.20 moles of Ethane will react with  =  X moles of Oxygen

Solving for X,

                            X  =  (4.20 mol × 7 mol) ÷ 2 mol

                            X  = 14.7 moles of Oxygen

Result:

           14.7 moles of Oxygen are required to react with 4.20 moles of Ethane for complete reaction.

a. the law of proportion states the combination of 2 atoms in a specific ratio to produce multiple compounds such as the given statement of question ‘1st compound contains 88.8 % cu ’ indicates 88.8 g of cu is present along with 11.2 g oxygen in 100 g of compound whereas the statement ‘2st compound contains 79.9 % cu ’ indicates 79.9 g of cu is present along with 20.1 g oxygen in 100 g of compound

1st compound contains 88.8 g cu

for cu

m = 63 g /mol

atomic % = 88.8 /63 = 1.42

for o

m = 16 g /mol

atomic % = 11.2 /16 = 0.7

b. the emperical formula of 1st compound is

cu1.4o0.7

cu2o