If 8.00 moles of NH3 reacted with 14.0 moles of O2, how many moles of H2O will be produced?
4NH3 (g) + 7O2 (g) --> 4NO2 + 6H2O (g)
How much H2O could you get if you used 8.00 moles NH3 and all of the oxygen needed. That will give you
8.00 x (6 moles H2O/4 moles NH3) = about 12.
Now how much H2O would be produced if you used 14.0 moles Oxygen and all of the NH3 needed. That will be 14.0 moles O2 x (6 moles H2O/7 moles O2) = 12
aha!. They are the same so you would get 12 moles H2O. What if they weren't the same. In that case you have a limiting reagent problem, the SMALLER value is the one you choose and the reagent producing the smaller value is the limiting reagent.
Well, let me calculate that for you. According to the balanced equation, 4 moles of NH3 will react with 7 moles of O2 to produce 6 moles of H2O. So, if we have 8.00 moles of NH3, it means we have an excess of ammonia because we only need 4 moles. Therefore, the maximum number of moles of H2O that can be produced is 6 moles. It's like having a buffet but only getting 6 plates.
To determine how many moles of H2O will be produced, we need to use the given mole ratios from the balanced chemical equation.
From the balanced equation, we can see that:
- 4 moles of NH3 react with 6 moles of H2O.
To find the number of moles of H2O produced, we can set up a proportion using the mole ratios:
(8 moles NH3) / (4 moles NH3) = (x moles H2O) / (6 moles H2O)
Cross-multiplying, we get:
8 moles NH3 * 6 moles H2O = 4 moles NH3 * x moles H2O
48 moles H2O = 4x
Dividing both sides by 4, we find:
12 moles H2O = x
Therefore, 12 moles of H2O will be produced when 8.00 moles of NH3 react with 14.0 moles of O2.
To determine the number of moles of H2O produced, we need to use the stoichiometry of the balanced chemical equation.
According to the balanced equation:
4NH3 (g) + 7O2 (g) --> 4NO2 + 6H2O (g)
The stoichiometric ratio between NH3 and H2O is 4:6, meaning that for every 4 moles of NH3 consumed, 6 moles of H2O are produced.
Given that we have 8.00 moles of NH3, we can set up a proportion to find the number of moles of H2O produced:
4 moles NH3 / 6 moles H2O = 8.00 moles NH3 / x moles H2O
Cross-multiplying, we get:
4 moles NH3 * x moles H2O = 8.00 moles NH3 * 6 moles H2O
Simplifying:
4x = 48
Dividing both sides by 4:
x = 12
Therefore, when 8.00 moles of NH3 react with 14.0 moles of O2, 12 moles of H2O will be produced.