3NO2(g) + H2O(l) → 2HNO3(l) + NO(g)
How many moles of NO2(g) are required to react with 1.00 mole of water to produce 3.00 moles of nitric acid?
You need 3 mols of NO2 for each mol of H2O
You can see that from your balanced equation.
However You get TWO mols of nitric acid, not 3
Well, let's break it down! According to the stoichiometry of the balanced equation, we see that 3 moles of NO2 are required to produce 2 moles of HNO3.
Since we want to produce 3 moles of HNO3, we can set up a proportion:
(3 moles NO2 / 2 moles HNO3) = (x moles NO2 / 1 mole H2O)
Cross-multiplying, we get:
3 moles NO2 = 2 moles HNO3 * 1 mole H2O / 1 mole NO2
Simplifying, we find:
3 moles NO2 = 2 moles HNO3
So, to produce 3 moles of HNO3, you would need 3 moles of NO2.
To determine the moles of NO2(g) required to react with 1.00 mole of water to produce 3.00 moles of nitric acid, we need to use the balanced equation:
3NO2(g) + H2O(l) → 2HNO3(l) + NO(g)
From the balanced equation, we can see that the stoichiometric ratio between NO2 and H2O is 3:1. This means that for every 3 moles of NO2, we need 1 mole of H2O.
To find the moles of NO2, we can set up a simple ratio:
3 moles NO2 / 1 mole H2O = x moles NO2 / 1 mole HNO3
Using the given information, we can plug in the values:
3.00 moles HNO3 = x moles NO2
Simplifying the equation, we get:
x = 3.00 moles NO2
Therefore, 3.00 moles of NO2(g) are required to react with 1.00 mole of water to produce 3.00 moles of nitric acid.
To determine the number of moles of NO2(g) required to react with 1.00 mole of water to produce 3.00 moles of nitric acid, we can use the balanced equation:
3NO2(g) + H2O(l) → 2HNO3(l) + NO(g)
According to the stoichiometry of the equation, we can see that the coefficient of NO2(g) is 3. This means that for every 3 moles of NO2(g), we will produce 2 moles of HNO3(l) and 1 mole of NO(g).
Since the ratio of moles in the reaction is 3:2:1 for NO2(g):HNO3(l):NO(g), we can set up a proportion to find the number of moles of NO2(g) required.
Let x represent the number of moles of NO2(g) required.
Using the proportion:
(3 moles NO2(g)) : (2 moles HNO3(l) + 1 mole NO(g)) = (x moles NO2(g)) : (3.00 moles HNO3(l) + 3.00 moles NO(g))
Cross-multiplying, we get:
3(x) = 2(3.00) + 1(3.00)
3x = 6.00 + 3.00
3x = 9.00
Dividing both sides by 3, we find:
x = 3.00 moles
Therefore, 3.00 moles of NO2(g) are required to react with 1.00 mole of water to produce 3.00 moles of nitric acid.