Sure, let's approach the problem step-by-step.
### 2.1.2 Calculate the number of moles of HCl (aq) added to the Sodium thiosulphate.
Given data:
- Molarity of HCl ([tex]\( M \)[/tex]) = 0.1 mol/L
- Volume of HCl ([tex]\( V \)[/tex]) = 0.025 L
To find the number of moles, we use the formula:
[tex]\[ \text{Number of moles} = \text{Molarity} \times \text{Volume} \][/tex]
Substituting the given values:
[tex]\[ \text{Number of moles of HCl} = 0.1 \, \text{mol/L} \times 0.025 \, \text{L} \][/tex]
[tex]\[ \text{Number of moles of HCl} = 0.0025 \, \text{moles} \][/tex]
Therefore, the number of moles of [tex]\( HCl \)[/tex] added is [tex]\( 0.0025 \)[/tex] moles.
### 2.1.3 Calculate the volume of [tex]\( SO_2(g) \)[/tex] that will be formed if the reaction takes place.
Given that the reaction between [tex]\(\text{HCl}\)[/tex] and Sodium thiosulphate goes to completion with a 1:1 stoichiometric ratio, the number of moles of [tex]\( SO_2 \)[/tex] produced will be the same as the number of moles of [tex]\( HCl \)[/tex] added, which is [tex]\( 0.0025 \)[/tex] moles.
To calculate the volume of [tex]\( SO_2 \)[/tex] gas formed, we can use the Ideal Gas Law:
[tex]\[ PV = nRT \][/tex]
Where:
- [tex]\( P \)[/tex] = Pressure (1 atm)
- [tex]\( V \)[/tex] = Volume of [tex]\( SO_2 \)[/tex] (in liters, which we need to find)
- [tex]\( n \)[/tex] = Number of moles of [tex]\( SO_2 \)[/tex] (0.0025 moles)
- [tex]\( R \)[/tex] = Ideal gas constant (0.0821 L·atm/(mol·K))
- [tex]\( T \)[/tex] = Temperature (298 K)
Rearranging the Ideal Gas Law to solve for [tex]\( V \)[/tex]:
[tex]\[ V = \frac{nRT}{P} \][/tex]
Substituting the given values:
[tex]\[ V = \frac{0.0025 \, \text{moles} \times 0.0821 \, \text{L·atm/(mol·K)} \times 298 \, \text{K}}{1 \, \text{atm}} \][/tex]
[tex]\[ V = \frac{0.0613675}{1} \][/tex]
[tex]\[ V = 0.061165 \, \text{L} \][/tex]
Therefore, the volume of [tex]\( SO_2 \)[/tex] that will be formed is approximately [tex]\( 0.061165 \)[/tex] liters.
