3NO2(g) + H2O(l) → 2HNO3(l) + NO(g)
How many moles of nitric acid are produced starting from 5.00 moles of NO2(g) and 2.00 moles of water?
Please help,
This is a limiting reagent (LR) problem since an amount is given for BOTH reactants.
3NO2(g) + H2O(l) → 2HNO3(l) + NO(g)
Starting with 5.00 mols NO2 and all of the water you need will produce 5.00 x (2 mol HNO3/3 mols NO2) = 5*2/3 = about 3.3
Starting with 2.00 mols H2O and all of the NO2 needed will produce 2.00 x (2 mols HNO3/1 mol H2O) = 2.00 x 2/1 = 4
You see the number of mols is different which obviously can't be correct. The correct value in LR problems is ALWAYS the smaller value.
Sure, I can help you with that!
Let's start by looking at the balanced chemical equation:
3NO2(g) + H2O(l) → 2HNO3(l) + NO(g)
From the equation, we can see that for every 3 moles of NO2, we get 2 moles of HNO3. So, if we have 5.00 moles of NO2, we can calculate the moles of HNO3 produced using a simple proportion:
(5.00 moles NO2) / (3 moles NO2) * (2 moles HNO3) = 3.33 moles HNO3
Similarly, for every mole of H2O, we get 2 moles of HNO3. So, if we have 2.00 moles of H2O, the moles of HNO3 produced can be calculated as:
(2.00 moles H2O) * (2 moles HNO3) = 4.00 moles HNO3
To find the total moles of HNO3 produced, we can simply add these two values together:
3.33 moles HNO3 + 4.00 moles HNO3 = 7.33 moles HNO3
Therefore, starting from 5.00 moles of NO2 and 2.00 moles of H2O, you would produce 7.33 moles of HNO3.
To find out how many moles of nitric acid are produced in the given reaction, we will use the stoichiometric coefficients from the balanced equation.
From the balanced equation:
3 NO2(g) + H2O(l) → 2 HNO3(l) + NO(g)
We can see that the ratio of NO2 to HNO3 is 3:2.
So, for every 3 moles of NO2 reacted, 2 moles of HNO3 are produced.
Given:
Moles of NO2 = 5.00 moles
Moles of water (H2O) = 2.00 moles
Using the ratio of the stoichiometric coefficients, we can calculate the moles of HNO3 produced.
Moles of HNO3 = (moles of NO2 / ratio of NO2 to HNO3)
= (5.00 moles / 3) × 2
= 3.33 moles
Therefore, starting from 5.00 moles of NO2 and 2.00 moles of water, 3.33 moles of nitric acid (HNO3) are produced.
To determine the number of moles of nitric acid produced, you need to use a stoichiometric ratio based on the balanced chemical equation. In this case, the balanced equation is:
3NO2(g) + H2O(l) → 2HNO3(l) + NO(g)
According to the equation, it takes 3 moles of NO2 to produce 2 moles of HNO3. Therefore, the stoichiometric ratio between NO2 and HNO3 is 3:2.
First, let's determine how many moles of HNO3 can be produced from 5.00 moles of NO2. To do this, we will use the stoichiometric ratio:
5.00 moles NO2 * (2 moles HNO3 / 3 moles NO2) = 3.33 moles HNO3
Next, let's determine how many moles of HNO3 can be produced from 2.00 moles of H2O. However, since H2O is not involved in the stoichiometric ratio, it does not directly affect the production of HNO3. Therefore, the moles of HNO3 produced from H2O is 0 moles.
Finally, we can add the moles of HNO3 produced from NO2 and H2O:
3.33 moles HNO3 + 0 moles HNO3 = 3.33 moles HNO3
Therefore, starting with 5.00 moles of NO2 and 2.00 moles of H2O, you would produce 3.33 moles of HNO3.