25 g of NH3 is mixed with 4 moles of O2 in the given reaction: 4NH3(g)+5O2(g) revers reaction 4NO(g)+6H2O(1). What mass of H2O formed?
We need to start by balancing the equation:
4NH3(g) + 5O2(g) → 4NO(g) + 6H2O(l)
Next, we need to determine which reactant is limiting, which will tell us how much water is formed. To do this, we'll use the mole ratio of NH3 to O2:
4 moles O2 × (4 moles NH3 / 5 moles O2) = 3.2 moles NH3
Since we only have 25 g of NH3, which is less than 3.2 moles, NH3 is the limiting reactant. This means that all of the NH3 will be used up and we'll have some O2 left over.
Using the mole ratio of NH3 to H2O, we can now calculate how much H2O is formed:
25 g NH3 × (1 mol NH3 / 17.03 g NH3) × (6 mol H2O / 4 mol NH3) × (18.02 g H2O / 1 mol H2O) = 45.5 g H2O
Therefore, 45.5 g of H2O is formed in the reaction.
To find the mass of H2O formed, we need to determine the limiting reactant first. The limiting reactant is the reactant that is completely consumed and determines the maximum amount of product that can be formed.
Given:
- 25 g of NH3
- 4 moles of O2
Let's calculate the moles of NH3 and O2:
Molar mass of NH3 (molecular weight of NH3) = 14.01 g/mol (nitrogen) + 3 * 1.01 g/mol (hydrogen) = 17.03 g/mol
Moles of NH3 = mass of NH3 / molar mass of NH3
= 25 g / 17.03 g/mol
≈ 1.47 moles
Therefore, we have 1.47 moles of NH3.
Moles of O2 = 4 moles (given)
Next, we need to determine the stoichiometric ratio between NH3 and H2O in the reaction:
From the balanced equation: 4NH3(g) + 6H2O(1)
The stoichiometric ratio between NH3 and H2O is 4:6, or simplified as 2:3.
Since we have 1.47 moles of NH3, we can calculate the maximum moles of H2O that can be formed:
Moles of H2O = (moles of NH3) x (3 moles of H2O / 2 moles of NH3)
= 1.47 moles x (3/2)
= 2.20 moles
Now, we need to calculate the mass of H2O formed:
Molar mass of H2O = 2 * 1.01 g/mol (hydrogen) + 16.00 g/mol (oxygen) = 18.02 g/mol
Mass of H2O = moles of H2O x molar mass of H2O
= 2.20 moles x 18.02 g/mol
≈ 39.64 g
Therefore, approximately 39.64 g of H2O is formed.