well, which is more?
2*5/4 ? 16/32
10/4 ? .5
2.5>.5
so H2 is far in excess.
5 grams of H2 reacts with 16 g of O2, which in excess?
2H2(g)+O2(g)=2H2O(g)
2 answers
Convert grams to moles, then divide moles by respective coefficients in balanced equation. The smaller value is the limiting reagent and the other(s) are in excess.The amounts used are proportional to the limiting reagent.
2H2 + O2 => 2H2O
moles H2 = 5g/ 2g/mol = 2.50 moles
moles O2 = 16g/32g/mol = 0.50 mole
for H2 divide moles by 2 => 2.5/2 => 1.25
for O2 divide moles by 1 => 0.5/1 => 0.50 (Limiting Reagent)
Since 0.50 < 1.25, O2 is the limiting reagent and H2 is in excess... Calculations of amount of H2 used is based upon the 0.50 mole O2. That is, moles H2 used is 2x moles of limiting reagent => 2(0.50mol)H2 used => 1.0 mole H2 used => Remaining H2 => 2.5 moles given - 1.0 moles used => 1.5 mole H2 remains in excess.
2H2 + O2 => 2H2O
moles H2 = 5g/ 2g/mol = 2.50 moles
moles O2 = 16g/32g/mol = 0.50 mole
for H2 divide moles by 2 => 2.5/2 => 1.25
for O2 divide moles by 1 => 0.5/1 => 0.50 (Limiting Reagent)
Since 0.50 < 1.25, O2 is the limiting reagent and H2 is in excess... Calculations of amount of H2 used is based upon the 0.50 mole O2. That is, moles H2 used is 2x moles of limiting reagent => 2(0.50mol)H2 used => 1.0 mole H2 used => Remaining H2 => 2.5 moles given - 1.0 moles used => 1.5 mole H2 remains in excess.
1 answer