What volume of .150 M potassium iodide solution will completely react with .155 L of .112 M lead (II) nitrate? equation: 2KI (aq) + Pb(NO3)2 (aq) ----> 2KNO3 (aq) + PbI2 (s)
(I'll use the balanced equation given in the question for this answer).
To solve this equation, let's first use the molarity equation to solve for the moles of lead(II) nitrate:
Now, we can use the coefficients of the chemical equation to calculate the relative number of moles of potassium iodide,
Now, let's use the molarity equation again to solve for the liters of the potassium iodide solution:
Therefore, you would need