Question

Calculate the mass of water produced when 7.63 g of butane reacts with excess oxygen.

Answer 1

You can calculate the mass of water by using the molar units to convert 7.63g of butane to moles and then find the molar mass of H2O which I think should be 1.01 + 1.01 + 16.00 if I remember correctly.  That is 18.02moles/g.  The butane is the limiting reactant but you would need the chemical equation to solve this

Answer 2

C₄H₁₀  +  $\frac{13}{2}$O₂   →   4CO₂  +  5H₂O

If mass of butane = 7.63 g
and mole = mass ÷ molar mass

then mole of butane = 7.63 g  ÷  ((12 × 4) + (1 × 10))
                                 = 7.63 g ÷ 58
                                 =  0.1316 mol

Mole Ratio of C₄H₁₀  :  H₂O     is    1    :    5

∴ if mole of butane =  0.1316 mol

then mole of water =  (0.1316 mol  ×   5 )
                               =  0.658 mol

Since mass = moles  ×  molar mass

then mass of water produced =  0.658 mol  ×  ((1 × 2) + (16 × 1))
                                                =  11.844 g