2.3g hydrogen react with 29g of oxygen to form water what is the limiting reagent
2H2 + O2 ==> 2H2O
mols H2 = 2.3/2 = 1.15
mols O2 = 29/32 = 0.906
Convert mols each to mols product.
1.15 x (2 mols H2O/2 mols H2) = 1.15
0.906 x (2 mols H2O/1 mol O2) = ?
The smaller amount of product is the product formed and the reagent responsible for that amount is the limiting reagent.
To determine the limiting reagent, we need to compare the number of moles of each reactant to the stoichiometric ratio of reactants in the balanced chemical equation.
First, convert the mass of each reactant to moles using the molar masses:
Molar mass of hydrogen (H₂) = 2.02 g/mol
Molar mass of oxygen (O₂) = 32.00 g/mol
Number of moles of hydrogen = mass of hydrogen / molar mass of hydrogen
= 2.3 g / 2.02 g/mol
≈ 1.14 mol
Number of moles of oxygen = mass of oxygen / molar mass of oxygen
= 29 g / 32.00 g/mol
≈ 0.91 mol
Next, we examine the balanced chemical equation for the reaction:
2H₂ + O₂ → 2H₂O
The stoichiometric ratio between hydrogen and oxygen is 2:1, which means that 2 moles of hydrogen react with 1 mole of oxygen.
Since the ratio of moles of hydrogen to moles of oxygen is 1.14:0.91, it is not in the same ratio as the stoichiometry. Therefore, oxygen is the limiting reagent because there is less of it available for reaction.
In conclusion, oxygen is the limiting reagent in this reaction.