<u>Answer:</u> The volume of barium hydroxide is 183 mL.
<u>Explanation:</u>
To calculate the moles of cadmium nitrate, we use the equation:
.....(1)
Molarity of
= 0.796 M
Volume of
= 161 mL = 0.161 L (Conversion factor: 1 L = 1000 mL)
Putting values in equation 1, we get:
![0.796mol/L=\frac{\text{Moles of }MnSO_4}{0.161L}\\\\\text{Moles of }MnSO_4=0.13mol](https://tex.z-dn.net/?f=0.796mol%2FL%3D%5Cfrac%7B%5Ctext%7BMoles%20of%20%7DMnSO_4%7D%7B0.161L%7D%5C%5C%5C%5C%5Ctext%7BMoles%20of%20%7DMnSO_4%3D0.13mol)
The chemical equation for the reaction of manganese sulfate and barium hydroxide follows:
![MnSO_4(aq.)+Ba(OH)_2(aq.)\rightarrow Mn(OH)_2(s)+BaSO_4(aq.)](https://tex.z-dn.net/?f=MnSO_4%28aq.%29%2BBa%28OH%29_2%28aq.%29%5Crightarrow%20Mn%28OH%29_2%28s%29%2BBaSO_4%28aq.%29)
By Stoichiometry of the reaction:
1 mole of manganese sulfate reacts with 1 mole of barium hydroxide.
So, 0.13 moles of manganese sulfate will react with =
of barium hydroxide
Now, calculating the volume of barium hydroxide by using equation 1, we get:
Moles of barium hydroxide = 0.13 moles
Molarity of barium hydroxide = 0.710 M
Putting values in equation 1, we get:
![0.710mol/L=\frac{0.13mol}{\text{Volume of barium hydroxide}}\\\\\text{Volume of barium hydroxide}=0.183L](https://tex.z-dn.net/?f=0.710mol%2FL%3D%5Cfrac%7B0.13mol%7D%7B%5Ctext%7BVolume%20of%20barium%20hydroxide%7D%7D%5C%5C%5C%5C%5Ctext%7BVolume%20of%20barium%20hydroxide%7D%3D0.183L)
Converting this into milliliters, we use the conversion factor:
1 L = 1000 mL
So, ![0.183L=0.183\times 1000=183mL](https://tex.z-dn.net/?f=0.183L%3D0.183%5Ctimes%201000%3D183mL)
Hence, the volume of barium hydroxide is 183 mL.