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.

1.consider the following balanced equation and answer parts a through. show calculations for each part.  

3 ag(s) + 4 hno3(aq) → 3 agno3(aq) + no(g) + 2 h2o(l)  

a. calculate the number of moles of agno3 produced by the reaction of 3 moles of ag with excess hno3. if the number to the left of the compound represents the number of moles of the compound which are present, then 3 moles of ag react with 4 moles hno3 to produce 3 moles agno3 (aq).

so the answer to a would be 3 moles.

b. calculate the number of moles of agno3 produced by the reaction of 108 grams of ag with excess hno3.  

i'm not sure how to plug in the grams of silver into an equation using moles to find how many moles are produced.

c. calculate the number of grams of agno3 produced by the reaction of 108 grams of ag with excess hno3  

d. calculate the number of moles of no produced as a byproduct of the reaction of 108 grams of ag with excess hno3.  

e. calculate the number of grams of no produced as a byproduct starting with 108 grams of ag and excess hno3.  

explanation: