To determine which reactant is in excess and by how many moles when 25.0 mL of 0.250 M HNO3 reacts with 70.0 mL of 0.150 M NaOH, we'll walk through the necessary steps to solve this problem.
### Step-by-Step Solution:
1. Find the number of moles of each reactant:
First, we convert the volumes of the solutions from milliliters (mL) to liters (L) since molarity (M) is defined as moles per liter.
- Volume of HNO3 in liters: [tex]\( \text{volume\_HNO3\_L} = \frac{25.0 \text{ mL}}{1000} = 0.025 \text{ L} \)[/tex]
- Volume of NaOH in liters: [tex]\( \text{volume\_NaOH\_L} = \frac{70.0 \text{ mL}}{1000} = 0.070 \text{ L} \)[/tex]
Next, we use the molarity and the volume to calculate the moles of each reactant.
- Moles of HNO3: [tex]\( \text{moles\_HNO3} = \text{molarity\_HNO3} \times \text{volume\_HNO3\_L} = 0.250 \text{ M} \times 0.025 \text{ L} = 0.00625 \text{ moles} \)[/tex]
- Moles of NaOH: [tex]\( \text{moles\_NaOH} = \text{molarity\_NaOH} \times \text{volume\_NaOH\_L} = 0.150 \text{ M} \times 0.070 \text{ L} = 0.0105 \text{ moles} \)[/tex]
2. Determine the limiting reactant:
According to the balanced chemical equation:
[tex]\[
\text{HNO3} + \text{NaOH} \rightarrow \text{NaNO3} + \text{H2O}
\][/tex]
The molar ratio between HNO3 and NaOH is 1:1.
To find the limiting reactant, compare the moles of HNO3 and NaOH:
- Moles of HNO3 = 0.00625
- Moles of NaOH = 0.0105
Since we need equal amounts of moles for a complete reaction and 0.00625 < 0.0105, HNO3 is the limiting reactant.
3. Determine the excess reactant and the amount in excess:
- Since HNO3 is the limiting reactant, all 0.00625 moles of HNO3 will react with an equal amount of NaOH.
- Therefore, the remaining moles of NaOH will be:
[tex]\[
\text{excess\_NaOH} = \text{moles\_NaOH} - \text{moles\_HNO3} = 0.0105 - 0.00625 = 0.00425 \text{ moles}
\][/tex]
### Conclusion:
The reactant in excess is NaOH, and the amount of NaOH in excess is 0.00425 moles.
