158.5 mL of a AgNO3 solution at 5.0M was combined with a 3.5M CaCl2 solution. 110.5g of AgCl was recovered. Given that the CaCl2 is the limiting reagent, how many milliliters of the 3.5M CaCl2 solution was used?
Write the equation and balance it. Convert 110.5 g AgCl to moles AgCl, convert to moles CaCl2, then M = moles/L, solve for L and convert to mL.
In order to find the number of milliliters of the 3.5M CaCl2 solution used, we need to use stoichiometry and the concept of limiting reagents.
To begin, we should first calculate the number of moles of AgNO3 that were used. We can use the given volume and molarity of the AgNO3 solution to find the amount in moles:
Moles of AgNO3 = Volume of solution (in liters) x Molarity of solution
= 158.5 mL x (1 L / 1000 mL) x 5.0 mol/L
= 0.7925 mol
Next, we can use the balanced chemical equation for the reaction between AgNO3 and CaCl2 to determine the stoichiometric ratio. The balanced equation is as follows:
2AgNO3 + CaCl2 -> 2AgCl + Ca(NO3)2
From the balanced equation, we can see that 2 moles of AgNO3 react with 1 mole of CaCl2 to produce 2 moles of AgCl.
Since CaCl2 is the limiting reagent, we need to determine the amount of CaCl2 that reacts with the AgNO3. Using the stoichiometric ratio from the balanced equation, we can calculate the moles of CaCl2 needed:
Moles of CaCl2 = (1/2) x Moles of AgNO3
= (1/2) x 0.7925 mol
= 0.39625 mol
Finally, we can determine the volume of the 3.5M CaCl2 solution used by rearranging the equation for molarity:
Volume of solution (in liters) = Moles of solute / Molarity of solution
= 0.39625 mol / 3.5 mol/L
= 0.113214 L
To convert the volume to milliliters, we multiply by 1000:
Volume of solution (in mL) = 0.113214 L x (1000 mL / 1 L)
≈ 113.21 mL
Therefore, approximately 113.21 milliliters of the 3.5M CaCl2 solution was used.