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? 2 sig figs
2AgNO3 + CaCl2 ==> 2AgCl + Ca(NO3)2
mole AgCl = 110.5/molar mass = ??
moles CaCl2 used = moles AgCl x (1 mole CaCl2/2 moles AgCl) = ??
M CaCl2 = moles CaCl2/L CaCl2.
Solve for L CaCl2 and convert to mL.
All of this depends, of course, upon the reaction being complete and recovery being 100% of the theoretical yield.
To find out how many milliliters of the 3.5M CaCl2 solution were used, we need to determine the number of moles of CaCl2 used in the reaction.
First, let's calculate the moles of AgCl formed using the mass of AgCl recovered. The molar mass of AgCl is 143.32 g/mol.
moles of AgCl = mass of AgCl / molar mass of AgCl
moles of AgCl = 110.5 g / 143.32 g/mol
moles of AgCl = 0.771 mol (rounded to 3 decimal places)
Since calcium chloride is the limiting reagent, the stoichiometry of the reaction tells us that 1 mole of AgCl is formed from 2 moles of CaCl2.
moles of CaCl2 = 0.771 mol / 2
moles of CaCl2 = 0.386 mol (rounded to 3 decimal places)
Now, we can use the molarity of the CaCl2 solution to calculate the volume used.
Molarity is defined as moles of solute per liter of solution. Since we want the volume in milliliters, we need to convert the molarity to moles per milliliter.
moles of CaCl2 per milliliter = molarity of CaCl2 solution / 1000
moles of CaCl2 per milliliter = 3.5 mol/L / 1000
moles of CaCl2 per milliliter = 0.0035 mol/mL (rounded to 4 decimal places)
Finally, we can calculate the volume (in milliliters) of the 3.5M CaCl2 solution used.
volume of CaCl2 solution = moles of CaCl2 / moles of CaCl2 per milliliter
volume of CaCl2 solution = 0.386 mol / 0.0035 mol/mL
volume of CaCl2 solution = 110.29 mL (rounded to 2 decimal places)
Therefore, approximately 110.29 mL (rounded to 2 decimal places) of the 3.5M CaCl2 solution was used.