Certainly! Let's go through the steps to determine how many moles of [tex]\( Ba(NO_3)_2 \)[/tex] could form given the quantities and concentrations of [tex]\( HNO_3 \)[/tex] and [tex]\( Ba(OH)_2 \)[/tex].
### Step-by-Step Solution:
#### 1. Calculation of Moles of Each Reactant:
- Moles of [tex]\( HNO_3 \)[/tex]:
- Given volume of [tex]\( HNO_3 \)[/tex]: [tex]\( 50.0 \)[/tex] mL
- Given molarity of [tex]\( HNO_3 \)[/tex]: [tex]\( 0.250 \)[/tex] M
- Convert volume from mL to L: [tex]\( 50.0 \, \text{mL} \times \frac{1 \, \text{L}}{1000 \, \text{mL}} = 0.0500 \, \text{L} \)[/tex]
- Calculate moles of [tex]\( HNO_3 \)[/tex]:
[tex]\[
\text{Moles of } HNO_3 = 0.250 \, \text{M} \times 0.0500 \, \text{L} = 0.0125 \, \text{moles}
\][/tex]
- Moles of [tex]\( Ba(OH)_2 \)[/tex]:
- Given volume of [tex]\( Ba(OH)_2 \)[/tex]: [tex]\( 30.0 \)[/tex] mL
- Given molarity of [tex]\( Ba(OH)_2 \)[/tex]: [tex]\( 0.400 \)[/tex] M
- Convert volume from mL to L: [tex]\( 30.0 \, \text{mL} \times \frac{1 \, \text{L}}{1000 \, \text{mL}} = 0.0300 \, \text{L} \)[/tex]
- Calculate moles of [tex]\( Ba(OH)_2 \)[/tex]:
[tex]\[
\text{Moles of } Ba(OH)_2 = 0.400 \, \text{M} \times 0.0300 \, \text{L} = 0.012 \, \text{moles}
\][/tex]
#### 2. Stoichiometry of the Reaction:
The balanced chemical equation is:
[tex]\[
2 \, HNO_3 + Ba(OH)_2 \rightarrow 2 \, H_2O + Ba(NO_3)_2
\][/tex]
- This equation tells us that 1 mole of [tex]\( Ba(OH)_2 \)[/tex] reacts with 2 moles of [tex]\( HNO_3 \)[/tex].
#### 3. Determine the Limiting Reagent:
- Calculate how many moles of [tex]\( HNO_3 \)[/tex] are needed to completely react with the available [tex]\( Ba(OH)_2 \)[/tex]:
[tex]\[
\text{Moles of } HNO_3 \text{ needed} = 2 \times \text{Moles of } Ba(OH)_2 = 2 \times 0.012 \, \text{moles} = 0.024 \, \text{moles}
\][/tex]
- Compare the moles of [tex]\( HNO_3 \)[/tex] needed with the moles of [tex]\( HNO_3 \)[/tex] available:
[tex]\[
\text{Available } HNO_3 = 0.0125 \, \text{moles}
\][/tex]
- Since [tex]\( 0.0125 \text{ moles} \)[/tex] of [tex]\( HNO_3 \)[/tex] is less than [tex]\( 0.024 \text{ moles} \)[/tex], [tex]\( HNO_3 \)[/tex] is the limiting reagent.
#### 4. Calculate Moles of [tex]\( Ba(NO_3)_2 \)[/tex] Formed:
- Since [tex]\( HNO_3 \)[/tex] is the limiting reagent, we use its moles to calculate the product:
- According to the balanced equation, 2 moles of [tex]\( HNO_3 \)[/tex] produce 1 mole of [tex]\( Ba(NO_3)_2 \)[/tex].
- Hence, moles of [tex]\( Ba(NO_3)_2 \)[/tex] formed:
[tex]\[
\text{Moles of } Ba(NO_3)_2 = \frac{0.0125 \, \text{moles of } HNO_3}{2} = 0.00625 \, \text{moles}
\][/tex]
Therefore, the number of moles of [tex]\( Ba(NO_3)_2 \)[/tex] that could form from the given quantities of [tex]\( Ba(OH)_2 \)[/tex] and [tex]\( HNO_3 \)[/tex] is [tex]\( \boxed{0.00625} \)[/tex] moles.
