3NO2(g) + H2O(l) → 2HNO3(l) + NO(g)
How many moles of the excess reactant remain if 2.00 moles of H2O(l) and 5.00 moles of NO2(g) are used?
Question 30 options:
1.67 mol H2O(l)
3.00 mol NO2(g)
4.00 mol NO2(g)
1.00 mol NO2(g)
0.33 mol H2O(l)
To find out how many moles of the excess reactant remain, we need to determine the limiting reactant first.
Step 1: Let's calculate the number of moles of HNO3 that can be produced from each reactant:
From 2.00 moles of H2O(l):
2.00 moles H2O(l) x (2 moles HNO3/1 mole H2O(l)) = 4.00 moles HNO3
From 5.00 moles of NO2(g):
5.00 moles NO2(g) x (2 moles HNO3/3 moles NO2(g)) = 3.33 moles HNO3
Step 2: Compare the moles of HNO3 produced from each reactant:
From the calculations, we can see that NO2(g) is the limiting reactant because it produces less HNO3 than H2O(l).
Step 3: Determine the amount of excess reactant remaining:
Since NO2(g) is the limiting reactant, we will calculate the moles of H2O(l) that reacted using the stoichiometry of the balanced equation.
From 5.00 moles NO2(g):
5.00 moles NO2(g) x (1 mole H2O(l)/3 moles NO2(g)) = 1.67 moles H2O(l)
Therefore, the answer is 1.67 moles of H2O(l) (Option: 1.67 mol H2O(l)).
To determine how many moles of the excess reactant remain, we first need to identify the limiting reactant.
1. Start by calculating the moles of products that can be formed from each reactant. For every 3 moles of NO2, 2 moles of HNO3 are formed and 1 mole of NO is formed.
From 5.00 moles of NO2:
- 5.00 moles of NO2 produce (5.00/3) * 2 = 3.33 moles of HNO3
- 5.00 moles of NO2 produce 5.00/3 = 1.67 moles of NO
From 2.00 moles of H2O:
- 2.00 moles of H2O cannot produce any moles of HNO3 or NO because it is not involved in the balanced equation.
2. Compare the moles of products that can be formed from each reactant. The reactant that produces the least amount of product is the limiting reactant. In this case, the limiting reactant is H2O because it does not produce any of the products.
3. Calculate how many moles of the excess reactant remain by subtracting the moles of the limiting reactant used from the initial moles of the excess reactant.
For NO2:
Initial moles - moles used = 5.00 mol - 0 mol = 5.00 mol
So, 5.00 moles of NO2 remain as the excess reactant.
Therefore, the correct answer is 4.00 mol NO2(g).