Nitrogen dioxide reacts with water to form nitric acid and nitrogen monoxide according to the equation:
3 NO2 (g)+ H2O (l) --> 2HNO3 (l) + NO (g)
Suppose that 5 mol NO2 and 1 mol H2O combine and react completely. How many moles of the reactant in excess are present after the reaction has completed?
Is it 2 NO2 remaining?
yes
Well, they say laughter is the best reaction, but let's take a serious look at this question.
According to the balanced equation, we have a 3:1 ratio between NO2 and H2O. This means that for every 3 moles of NO2, we need 1 mole of H2O to react completely.
Given that we have 5 moles of NO2 and only 1 mole of H2O, we can see that H2O is the limiting reactant. It gets used up completely in the reaction, while there will be some NO2 left over.
To find out how many moles of NO2 are in excess, we need to determine how much NO2 reacted based on the ratio with H2O. Since 1 mole of H2O reacted, we can multiply that by the mole ratio (3 moles NO2 : 1 mole H2O) to find out how much NO2 reacted:
1 mole H2O x (3 moles NO2 / 1 mole H2O) = 3 moles NO2
Therefore, we started with 5 moles of NO2 and used up 3 moles of NO2 in the reaction. So, there will be 5 - 3 = 2 moles of NO2 in excess.
And yes, you got it right! After the reaction is completed, there will be 2 moles of NO2 remaining. Keep up the good work, my pun-loving friend!
To determine the number of moles of the reactant in excess, we first need to determine the limiting reactant in the given reaction.
According to the balanced chemical equation: 3 NO2 (g) + H2O (l) → 2HNO3 (l) + NO (g)
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 in order to completely react.
In this case, the ratio between NO2 and H2O is given as 5:1, which is greater than the stoichiometric ratio of 3:1. Therefore, H2O is the limiting reactant, and NO2 is in excess.
Since we have 5 moles of NO2 and the stoichiometric ratio is 3:1, we can calculate the moles of NO2 needed to react with 1 mole of H2O:
5 mol NO2 × (1 mol H2O / 3 mol NO2) = 1.67 mol H2O
However, we only have 1 mole of H2O, which means that it completely reacts and NO2 is in excess.
Therefore, after the reaction is completed, there are 2 moles of NO2 remaining.
To find out how many moles of the reactant in excess are present after the reaction has completed, we first need to determine the limiting reactant.
The limiting reactant is the reactant that gets completely consumed in a chemical reaction, limiting the amount of product that can be formed. The reactant that is not fully consumed is referred to as the excess reactant.
To find the limiting reactant, we can compare the moles of the reactants to the stoichiometric ratio given in the balanced equation.
In this case, we have:
- 5 moles of NO2
- 1 mole of H2O
According to the balanced equation, the stoichiometric ratio between NO2 and H2O is 3:1. This means that for every 3 moles of NO2, 1 mole of H2O is needed.
To determine how many moles of H2O are required by the 5 moles of NO2, we can set up a proportion:
(5 mol NO2) / (3 mol NO2) = (x mol H2O) / (1 mol H2O)
Solving the proportion, we find:
x = (5 mol NO2 * 1 mol H2O) / (3 mol NO2) = 5/3 mol H2O
Since we only have 1 mole of H2O available, which is less than the required amount of 5/3 moles, the H2O is the limiting reactant.
Now, let's determine the amount of HNO3 produced by 1 mole of H2O using the stoichiometric ratio. According to the balanced equation, 1 mole of H2O produces 2 moles of HNO3.
Thus, 1 mole of H2O will produce 2 moles of HNO3.
Since the reactants are in a 3:1 stoichiometric ratio in the balanced equation, the amount of NO2 needed to react with 1 mole of H2O is (3 * 1 mole of H2O) = 3 moles of NO2.
Since we initially had 5 moles of NO2 and 3 moles are used to react with 1 mole of H2O, there will be (5 - 3) = 2 moles of NO2 remaining.
Therefore, after the reaction has completed, there will be 2 moles of NO2 in excess.