5 grams of H2 reacts with 16 g of O2, which in excess?
2H2(g)+O2(g)=2H2O(g)
well, which is more?
2*5/4 ? 16/32
10/4 ? .5
2.5>.5
so H2 is far in excess.
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.
To determine which reactant is in excess, we need to compare the stoichiometry of the reactants and see which one is not fully consumed.
First, let's calculate the number of moles for each reactant using their molar masses:
Molar mass of H2: 2 g/mol
5 g of H2 = 5 g / 2 g/mol = 2.5 mol of H2
Molar mass of O2: 32 g/mol
16 g of O2 = 16 g / 32 g/mol = 0.5 mol of O2
From the balanced chemical equation, we can see that the stoichiometric ratio between H2 and O2 is 2:1. This means that 2 moles of H2 react with 1 mole of O2.
Since we have 2.5 moles of H2 and only 0.5 moles of O2, it is clear that there is excess H2 present. Therefore, H2 is in excess, while O2 is the limiting reactant.
To summarize:
- H2 is in excess because there are more moles of H2 than needed according to the stoichiometry.
- O2 is the limiting reactant because it will be completely consumed, leaving some unreacted H2 after the reaction.