What volume of 0.55M ... ?

Question

What volume of 0.55M nickel (II) nitrate, Ni(NO3)2, will react with 85mL of 0.25M potassium carbonate, K2CO3, to form nickel (II) carbonate, NiCO3, and potassium nitrate, KNO3?

How would I go about solving the problem? Thank you.

Answer 1

To solve the problem, you have to think systematically. In this problem, you know the volume and the concentration of the potassium carbonate solution and you are after the volume of the nickel solution.

First you need to write out a balanced chemical equation for the reaction.

Then, taking the molar concentration (0.25 M) and multiplying that by the volume of that solution in Liters (0.085L) gives you the number of moles of potassium carbonate that you are beginning with.

From the balanced chemical equation for this reaction, you see that 1 mole of potassium carbonate will react with 1 mole of nickel (II) nitrate, so you will need as many moles of nickel (II) nitrate as you have moles of potassium carbonate.

Now, if you divide that number of moles by the molar concentration of the nickel solution, you will find the volume (in Liters) of the nickel solution that you will need.

Answer 2

first, write the balanced equation:

Ni(NO3)2 + K2(CO3) >> Ni(CO3) + 2K(NO3)

then solve for volume of Nickel II Nitrate:

first (85 mL x .25 mol)/1000ml = 0.2125 moles of K2(CO3)

molarity is moles/liters
0.25M=0.25moles/1L

since the ratio between Ni(NO3) and K2(CO3) is 1 to 1, 0.2125 moles of K2(CO3) x 22.4 L = Liters of Ni(NO3) = 4.76L

So 4.76L is the volume of the reactant.