the product is water (H2O)
the reaction ratio is by moles ... 2 moles H to 1 mole O
convert the grams to moles (divide by atomic mass)
1.5g of H reacts with14.5g Oxygen
1)which is the limiting reagent?
2)calculate the amount of reactant which remains unreacted?
2 answers
2H2 + O2 ==> 2H2O
mols H2 = gram/molar mass = 1.5/2 = 0.75
mols O2 = 14.5/32 = approx 0.45 but you need to do these values again and more accurately. All of my calculations are estimates.Convert H2 to mols H2O produced IF you had all of the oxygen needed (an excess of oxygen). That is 0.75 mols H2 x (2 mols H2O/2 mols H2) = 0.75 mols H2O produced.
Now convert 0.45 mols O2 to mols H2O produced IF you had all of the H2 needed (an excess of H2). That is 0.45 x (2 mols H2O/1 mol O2) = 0.9 mols H2O produced.
Compare the two values. In limiting reagent (LR) problems the smaller number is always the correct one; therefore, the reaction will produce 0.75 mols H2O and H2 is the LR. O2 is the excess reagent (ER). How much O2 is used in the reaction. That is 0.75 mols H2 x (1 mol O2/2 mols H2) = about 0.375 mols O2. So how much is left? That is 0.45 initially - 0.375 mols used = ? mols O2 remaining un-reacted. Post your work if you get stuck.
mols H2 = gram/molar mass = 1.5/2 = 0.75
mols O2 = 14.5/32 = approx 0.45 but you need to do these values again and more accurately. All of my calculations are estimates.Convert H2 to mols H2O produced IF you had all of the oxygen needed (an excess of oxygen). That is 0.75 mols H2 x (2 mols H2O/2 mols H2) = 0.75 mols H2O produced.
Now convert 0.45 mols O2 to mols H2O produced IF you had all of the H2 needed (an excess of H2). That is 0.45 x (2 mols H2O/1 mol O2) = 0.9 mols H2O produced.
Compare the two values. In limiting reagent (LR) problems the smaller number is always the correct one; therefore, the reaction will produce 0.75 mols H2O and H2 is the LR. O2 is the excess reagent (ER). How much O2 is used in the reaction. That is 0.75 mols H2 x (1 mol O2/2 mols H2) = about 0.375 mols O2. So how much is left? That is 0.45 initially - 0.375 mols used = ? mols O2 remaining un-reacted. Post your work if you get stuck.