If 4.0 g of hydrogen and 10.0 g of oxygen are mixed according to the equation 2 H2 + 02 = 2 H20, which is the limiting reagent?
1. none, they are evenly mixed
2. oxygen
3. hydrogen
4. water
Figure the moles of each. 4gH2 is 2 moles about, while 10g O2 is about a 1/3 mole. I suspect O2 is limiting.
To determine the limiting reagent in a chemical reaction, we need to compare the number of moles of each reactant to the stoichiometric ratio in the balanced equation.
First, we need to find the number of moles for each reactant. We can do this by dividing the mass of each reactant by its molar mass.
The molar mass of hydrogen (H2) is 2.016 g/mol, and the molar mass of oxygen (O2) is 32.00 g/mol.
For hydrogen:
moles of hydrogen = mass of hydrogen / molar mass of hydrogen
moles of hydrogen = 4.0 g / 2.016 g/mol
moles of hydrogen ≈ 1.987 mol
For oxygen:
moles of oxygen = mass of oxygen / molar mass of oxygen
moles of oxygen = 10.0 g / 32.00 g/mol
moles of oxygen ≈ 0.3125 mol
Now, let's compare the moles of each reactant to the stoichiometric ratio in the balanced equation. The balanced equation tells us that for every 2 moles of hydrogen (H2), we need 1 mole of oxygen (O2).
From the calculations above, we have:
moles of hydrogen = 1.987 mol
moles of oxygen = 0.3125 mol
Comparing the ratio of moles, we find that the ratio of hydrogen to oxygen is approximately 6.35:1. Therefore, there is an excess of hydrogen.
So, the limiting reagent is oxygen because it is present in a smaller amount compared to the stoichiometric ratio. The limiting reagent determines the maximum amount of product that can be formed.