a mixture of 4.00 moles of hydrogen and
3.00 moles of oxygen is ignited, forming water. what is the maximum amount of water in grams that can be formed?
H2 + O2 -> H2O
balancing this,
2H2 + O2 -> 2H2O
then we find the limiting reactant. To determine the limiting reactant, we get the moles of product from the given moles of reactant. Whichever yield less number of moles of product is the limiting reactant. thus,
4 mol H2 * (2 mol H2O / 2 mol H2) = 4 mol H2O
3 mol O2 * (2 mol H2O / 1 mol O2) = 6 mol H2O
thus, the limiting reactant is H2.
finally, to get the mass of water, we use the moles of H2O produced by H2O, and multiply this by the molecular mass of water. Note that the molecular mass of H2O = 18 g/mol. thus,
4 * 18 = 72 g H2O
To determine the maximum amount of water that can be formed, we first need to understand the balanced chemical equation for the reaction between hydrogen and oxygen to form water. The balanced equation is:
2H₂ + O₂ → 2H₂O
From this balanced chemical equation, we can see that 2 moles of hydrogen react with 1 mole of oxygen to produce 2 moles of water.
Given that the mixture consists of 4.00 moles of hydrogen and 3.00 moles of oxygen, we need to determine which reactant is the limiting reactant. The limiting reactant is the one that is completely consumed and determines the maximum amount of product that can be formed.
To find the limiting reactant, we compare the number of moles of each reactant to the stoichiometry of the balanced equation.
For hydrogen:
Given: 4.00 moles
Stoichiometry: 2 moles of hydrogen react with 1 mole of oxygen
For oxygen:
Given: 3.00 moles
Stoichiometry: 1 mole of oxygen reacts with 2 moles of hydrogen
Therefore, the ratio of moles of hydrogen to oxygen is 4.00:3.00 for the respective stoichiometry.
From this ratio, we can see that there is an excess of hydrogen. This means that oxygen is the limiting reactant because there is not enough oxygen to completely react with the hydrogen.
To calculate the maximum amount of water that can be formed, we use the stoichiometry of the balanced equation. Since the ratio is 2:1 for hydrogen to water, we know that 1 mole of oxygen will react with 2 moles of hydrogen to produce 2 moles of water.
So, 3.00 moles of oxygen will produce 2.00 x (3.00 mol) = 6.00 moles of water.
To convert moles to grams, we need to use the molar mass of water, which is approximately 18.02 g/mol.
Therefore, the maximum amount of water in grams that can be formed is:
6.00 moles x 18.02 g/mol = 108.12 grams