Question

How many grams of water can be produced when 65.5 grams of sodium hydroxide reacts with excess sulfuric acid?

Answer 1

5°/. 2 C2H6 + 7 O2 → 4 CO2 + 6 H2O

M(H2O) = 2M(H) + M(O) = (2 x 1) + 16 = 18 g/mol


From the equation :

2 mol of C2H6 -----> 6 mol of H2O
11,7 mol of C2H6 -----> (6 x 11,7) / 2 = 35,1 mol of H2O

m(H2O) = M(H2O) x n(H2O)

m(H2O) = 18 x 35,1 = 631,8 g of H2O