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.
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?
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