This is a limiting reagent problem. You know that because amounts are given for BOTH reactants.
2H2 + O2 ==> 2H2O
mols H2 = grams/molar mass = ?
mols O2 = grams/molar mass = ?
Using the coefficients in the balanced equation, convert mols H2 to mols H2O. Do the same for mols O2 to mols H2O.
It is likely that the two values for mols H2O will not agree which means one of them is not right; the correct value in limiting reagent problems is ALWAYS the smaller value and the reagent producing that value is the limiting regent.
Now convert mols H2O to g. g = mols x molar mass.
How much water would be formed if 2.4 grams of hydrogen reacted with 19.3 grams of oxygen?
3 answers
Using the molar masses of hydrogen and oxygen, you can discover your limiting reactant (the reactant that would produce the least water). The water produced by your limiting reactant would be the amount of water formed.
(2.4 g H)(1 mol H/1.008 g)(1 mol H20/2 mol H)=1.19 mol H20
(19.3 g O)(1 mol O/16 g)(1 mol H20/i mol O)=1.21 mol H2O
So hydrogen would be your limiting reactant. To find grams of water, you'd use the molar mass again.
(1.19 mol H2O)(18.016 g/1 mol H2O)=21.4 g H20
(2.4 g H)(1 mol H/1.008 g)(1 mol H20/2 mol H)=1.19 mol H20
(19.3 g O)(1 mol O/16 g)(1 mol H20/i mol O)=1.21 mol H2O
So hydrogen would be your limiting reactant. To find grams of water, you'd use the molar mass again.
(1.19 mol H2O)(18.016 g/1 mol H2O)=21.4 g H20
*To be entirely correct, you'll want to use the values for hydrogen and oxygen gas. I'll let you figure those out, along with the molar ratios for hydrogen gas and water and oxygen and water.