if 8.00 mol of NH3reacted with O2 how many moles of H2O will be produced?
4NH3(g) + 7O2(g) -> 4NO2 + 6H2O(g)
1 year ago
1 year ago
To determine the number of moles of H2O produced when 8.00 mol of NH3 reacts with O2, we can use the balanced chemical equation provided.
The balanced chemical equation is:
4NH3(g) + 7O2(g) -> 4NO2(g) + 6H2O(g)
From the equation, we can see that 4 moles of NH3 react to produce 6 moles of H2O.
Therefore, we can set up a ratio to determine the number of moles of H2O produced:
4 moles NH3 : 6 moles H2O
We have 8.00 moles of NH3, so we can set up the following proportion:
4 moles NH3 / 8.00 moles NH3 = 6 moles H2O / x
To find x (the number of moles of H2O produced), we can cross-multiply and solve for x:
4 * x = 8.00 * 6
4x = 48.00
x = 48.00 / 4
x = 12.00
Therefore, when 8.00 mol of NH3 reacts with O2, 12.00 moles of H2O will be produced.
11 months ago
To find out how many moles of H2O will be produced when 8.00 mol of NH3 reacts with O2, we need to calculate it using the stoichiometry of the balanced chemical equation.
From the balanced equation: 4NH3(g) + 7O2(g) -> 4NO2 + 6H2O(g)
We can see that for every 4 moles of NH3 that react, 6 moles of H2O are produced.
So to find out how many moles of H2O will be produced, we need to use a proportion based on the stoichiometric ratio:
(6 moles H2O / 4 moles NH3) * 8.00 moles NH3 = 12.00 moles H2O
Therefore, 8.00 moles of NH3 will produce 12.00 moles of H2O.