In this type of question firstly we have to find the limiting reagent (means the reactant which is havimg less moles than the other reactant )

like for c2h6- moles = given mass of c2h6 ÷( molar mass × stochiometric coficient)

moles of c2h6 = 15÷(30×2)

= 0.25

moles of o2 = 45 ÷ ( 16 ×7)

= 0.40

so limting reagent is c2h6

now , moles of c2h6 is equal to the moles of water

0.25= mass of water produced ÷ ( molar mass of water × stochiometric coff.)

0.25 = m ÷ ( 18 × 6)

m = 27g

Explanation:

RATE MINE 5 STAR