If 4.0 g of hydrogen and 10.0 g of oxygen are mixed according to the equation 2 H2 + O2 = 2 H2O, which is the limiting reagent?
The one that produces fewer moles H2O.
Calculate each by itself and see which does that.
To determine the limiting reagent, we need to compare the moles of hydrogen and oxygen in the given amounts.
First, we need to find the number of moles of each element:
1. Hydrogen (H2):
Given weight = 4.0 g
Molar mass of H2 = 2.02 g/mol
Number of moles = given weight / molar mass = 4.0 g / 2.02 g/mol = 1.98 mol (approximately)
2. Oxygen (O2):
Given weight = 10.0 g
Molar mass of O2 = 32.00 g/mol
Number of moles = given weight / molar mass = 10.0 g / 32.00 g/mol = 0.31 mol (approximately)
Next, we need to compare the ratio of moles for both hydrogen and oxygen, based on the balanced chemical equation: 2 H2 + O2 = 2 H2O.
According to the equation, 2 moles of hydrogen react with 1 mole of oxygen to produce 2 moles of water.
Now, let's check the mole ratio:
Hydrogen (H2) to Oxygen (O2) ratio based on the equation is 2:1.
From the calculations, we found:
- Hydrogen (H2) has 1.98 moles
- Oxygen (O2) has 0.31 moles
To determine the limiting reagent, we need to compare the mole ratio to the actual number of moles present.
In this case, the mole ratio of H2 to O2 is 2:1. Since the actual number of moles of oxygen (0.31 mol) is less than the required number of moles (0.62 mol) based on the mole ratio, oxygen is the limiting reagent.
Therefore, oxygen is the limiting reagent in this reaction.