Calculate the number of milliliters of 0.710 m ba(oh)2 required to precipitate all of the mn2+ ions in 161 ml of 0.796 m kmno4 solution as mn(oh)2. the equation for the reaction is: mnso4(aq) + ba(oh)2(aq) mn(oh)2(s) + baso4(aq)

The volume of barium hydroxide is 183 mL.

Explanation:

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:

The chemical equation for the reaction of manganese sulfate and barium hydroxide follows:

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:

Converting this into milliliters, we use the conversion factor:

1 L = 1000 mL

So,

Hence, the volume of barium hydroxide is 183 mL.

i believe the answer is a.