Let's solve this problem step-by-step to find the heat change for the reaction.
### Step 1: Calculate the moles of [tex]\( \text{HNO}_3 \)[/tex] and [tex]\( \text{Ba(OH)}_2 \)[/tex].
Given:
- Volume of [tex]\( \text{HNO}_3 \)[/tex] solution: [tex]\( V_{\text{HNO}_3} = 50.0 \)[/tex] mL
- Concentration of [tex]\( \text{HNO}_3 \)[/tex] solution: [tex]\( M_{\text{HNO}_3} = 0.25 \)[/tex] M
Convert the volume to liters:
[tex]\[ \text{HNO}_3 \, \text{(moles)} = V_{\text{HNO}_3} \times M_{\text{HNO}_3} = 50.0 \, \text{mL} \times \frac{0.25 \, \text{mol}}{1000 \, \text{mL}} = 0.0125 \, \text{moles} \][/tex]
Given:
- Volume of [tex]\( \text{Ba(OH)}_2 \)[/tex] solution: [tex]\( V_{\text{Ba(OH)}_2} = 30.0 \)[/tex] mL
- Concentration of [tex]\( \text{Ba(OH)}_2 \)[/tex] solution: [tex]\( M_{\text{Ba(OH)}_2} = 0.40 \)[/tex] M
Convert the volume to liters:
[tex]\[ \text{Ba(OH)}_2 \, \text{(moles)} = V_{\text{Ba(OH)}_2} \times M_{\text{Ba(OH)}_2} = 30.0 \, \text{mL} \times \frac{0.40 \, \text{mol}}{1000 \, \text{mL}} = 0.012 \, \text{moles} \][/tex]
### Step 2: Determine the limiting reactant.
Reaction:
[tex]\[ 2 \text{HNO}_3 + \text{Ba(OH)}_2 \rightarrow 2 \text{H}_2\text{O} + \text{Ba(NO}_3)_2 \][/tex]
Moles ratio [tex]\( \text{HNO}_3 \)[/tex]: [tex]\( \text{Ba(OH)}_2 \)[/tex] is 2:1.
To react with [tex]\( 0.012 \)[/tex] moles of [tex]\( \text{Ba(OH)}_2 \)[/tex]:
The [tex]\( \text{HNO}_3 \)[/tex] needed:
[tex]\[ \text{HNO}_3 \, \text{needed} = 2 \times 0.012 \, \text{moles} = 0.024 \, \text{moles} \][/tex]
Since only 0.0125 moles of [tex]\( \text{HNO}_3 \)[/tex] are available and it's less than 0.024 moles, [tex]\( \text{HNO}_3 \)[/tex] becomes the limiting reactant.
Calculate limiting reactant in terms of equivalence:
[tex]\[ \frac{0.0125}{2} = 0.00625 \, \text{equivalents} \][/tex]
Hence, the limiting reactant is 0.00625 moles (equivalents).
### Step 3: Calculate the total mass of the solution.
Volume:
[tex]\[ \text{Total Volume} = 50.0 \, \text{mL} + 30.0 \, \text{mL} = 80.0 \, \text{mL} \][/tex]
Density: [tex]\( d_{\text{soln}} = 1.05 \, \text{g/mL} \)[/tex]
Total mass:
[tex]\[ \text{Total Mass} = 80.0 \, \text{mL} \times 1.05 \, \text{g/mL} = 84.0 \, \text{g} \][/tex]
### Step 4: Calculate the heat absorbed by the solution.
Specific Heat Capacity of the solution: [tex]\( c_{\text{soln}} = 4.2 \, \text{J/g°C} \)[/tex]
Temperature Change: [tex]\( \Delta T = 1.2 \, \text{°C} \)[/tex]
Heat absorbed by the solution:
[tex]\[ q_{\text{soln}} = \text{Total Mass} \times c_{\text{soln}} \times \Delta T = 84.0 \, \text{g} \times 4.2 \, \text{J/g°C} \times 1.2 \, \text{°C} = 423.36 \, \text{J} \][/tex]
### Step 5: Calculate the heat absorbed by the calorimeter.
Specific Heat Capacity of the calorimeter: [tex]\( c_{\text{cal}} = 4.5 \, \text{J/°C} \)[/tex]
Heat absorbed by the calorimeter:
[tex]\[ q_{\text{cal}} = c_{\text{cal}} \times \Delta T = 4.5 \, \text{J/°C} \times 1.2 \, \text{°C} = 5.4 \, \text{J} \][/tex]
### Step 6: Calculate the total heat change.
[tex]\[ q_{\text{total}} = q_{\text{soln}} + q_{\text{cal}} = 423.36 \, \text{J} + 5.4 \, \text{J} = 428.76 \, \text{J} \][/tex]
### Final Answer:
The heat change for the reaction is [tex]\( 428.76 \, \text{J} \)[/tex].
