How much water would be formed if 3.00 mol NH3 reacted with excess oxygen gas in the following equation?
Given the reaction 4NH3+ 5O2 --> 4NO + 6H2O
A) 2.00 mol
B) 3.00 mol
C) 4.50 mol
D) 6.00 mol
E) None of these
To determine how much water would be formed when 3.00 mol of NH3 reacts with excess oxygen gas in the given equation, we need to use the stoichiometry of the reaction.
Looking at the balanced equation, we can see that the ratio of NH3 to H2O is 4:6. This means that for every 4 moles of NH3, we would expect to produce 6 moles of H2O.
Since we have 3.00 moles of NH3, we can set up a proportion to find the amount of water formed:
(3.00 mol NH3) / (4 mol NH3) = (x mol H2O) / (6 mol H2O)
Simplifying the equation:
3.00 / 4 = x / 6
Cross-multiplying:
3.00 * 6 = 4 * x
18.00 = 4x
Dividing both sides by 4:
18.00 / 4 = x
4.50 = x
Therefore, the amount of water formed when 3.00 mol of NH3 reacts is 4.50 mol.
The correct answer is C) 4.50 mol.
3/4=6/x
E.