Answer:
a)  molar composition of this gas on both a wet and a dry basis are 
5.76 moles and 5.20 moles respectively.
Ratio of moles of water to the moles of dry gas =0.108 moles
b) Total air required = 68.51 kmoles/h
So, if combustion is 75% complete; then it is termed as incomplete combustion which require the same amount the same amount of air but varying product will be produced.
Explanation:
Let assume we have 100 g of mixture of gas:
Given that :
Mass of methane =75 g
Mass of ethane = 10 g
Mass of ethylene = 5 g
∴ Mass of the balanced water: 100 g - (75 g + 10 g + 5 g)
Their molar composition can be calculated as follows:
Molar mass of methane 
Molar mass of ethane 
Molar mass of ethylene 
Molar mass of water 
number of moles = 
Their molar composition can be calculated as follows:

 4.69 moles
 4.69 moles 

 0.33 moles
 0.33 moles

 0.18 moles
 0.18 moles

 0.56 moles
 0.56 moles
Total moles of gases for wet basis = (4.69 + 0.33 + 0.18 + 0.56) moles
= 5.76 moles
Total moles of gas for dry basis = (5.76 - 0.56)moles
= 5.20 moles
Ratio of moles of water to the moles of dry gas = 
= 
= 0.108 moles 
b) If 100 kg/h of this fuel is burned with 30% excess air(combustion); then we have the following equations:
     
4.69         2× 4.69 
moles       moles 
    
0.33      3.5 × 0.33
moles    moles
     
0.18           3× 0.18 
moles        moles
Mass flow rate = 100 kg/h
Their Molar Flow rate is as follows; 

Total moles of  required = (2 × 4.69) + (3.5 × 0.33) + (3 × 0.18) k moles
 required = (2 × 4.69) + (3.5 × 0.33) + (3 × 0.18) k moles
 = 11.075 k moles.
In 1 mole air = 0.21 moles 
Thus, moles of air required =  
 
= 52.7 k mole
30% excess air = 0.3 × 52.7 k moles 
= 15.81 k moles
Total air required = (52.7 + 15.81 ) k moles/h
 = 68.51 k moles/h
So, if combustion is 75% complete; then it is termed as incomplete combustion which require the same amount the same amount of air but varying product will be produced.