Mass of water produced when 9.69 g of butane reacts with excess oxygen.

C₄H₁₀  +  O₂   →   4CO₂  +  5H₂O

If mass of butane = 9.69 g
and mole = mass ÷ molar mass

then mole of butane = 9.69 g  ÷  ((12 × 4) + (1 × 10))
                                 = 9.69 g ÷ 58
                                 =  0.1671 mol

Mole Ratio of C₄H₁₀  :  H₂O     is    1    :    5

∴ if mole of butane =  0.1671 mol

then mole of water =  (0.1671 mol  ×   5 )
                               =  0.8353 mol

Since mass = moles  ×  molar mass

then mass of water produced =  0.8353 mol  ×  ((1 × 2) + (16 × 1))
                                                  =  15.036 g