Respuesta :
We will do stoichiometry in this kind of problem:
8 g H2 x (1 mol H2 / 2.0 g H2) x (2 mol H2O / 2 mol H2) x (18 g H20 / 1 mol H2O) = 72 g H2O
64 g O2 x (1 mol O2 / 32 g O2) x (2 mol H2O / 1 mol O2) x (16 g H2O / 1 mol H2O) = 72 g H2O
Therefore, the total mass of water formed is 72 g.
8 g H2 x (1 mol H2 / 2.0 g H2) x (2 mol H2O / 2 mol H2) x (18 g H20 / 1 mol H2O) = 72 g H2O
64 g O2 x (1 mol O2 / 32 g O2) x (2 mol H2O / 1 mol O2) x (16 g H2O / 1 mol H2O) = 72 g H2O
Therefore, the total mass of water formed is 72 g.