How many moles of water are produced when 1.05 g of oxygen gas reacts with 1.22 g of hydrogen gas to produce water?

Answer 1

1.05 g of O2 is 1.05/32 = 0.0328125 moles

1.22 g of H2 is 1.22/2.016 = 0.60516 moles

O2 is the limiting reactant. There is more than enough H2 to combine with the available oxygen. Each mole of O2 forms 2 moles of H2O, according to the reaction

2H2 = O2 -> 2H2O

Therefore, double 0.0328125 moles
to get the answer