How many grams of silver chloride can be prepared by the reaction of 122.2 mL of 0.22 M silver nitrate with 122.2 mL of 0.16 M calcium chloride?
Balance the equation. Find the moles of each reactant, compare them.
Balance:
2AgNO3+CaCl2 >> 2AgCl(s) + Ca(NO3)2
Note that you use 2 moles silver nitrate to each one mole of calcium chloride.
Moles each:
AgNO3: .1222*.22=.027moles
CaCl2: .1222*.16=.020 moles
so, you do not have twice as much of silver nitrate, so it is the limiting reactant.
You will use in the reaction .027moles silver nitrate, and .027/2 moles of CaCl2
moles of product: .027moles. Convert that to grams.
To determine the number of grams of silver chloride that can be prepared, we need to use the principles of stoichiometry.
First, let's write a balanced chemical equation for the reaction between silver nitrate (AgNO3) and calcium chloride (CaCl2):
2 AgNO3 + CaCl2 -> 2 AgCl + 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. Therefore, the molar ratio between AgNO3 and AgCl is 2:2, or 1:1.
Now, let's calculate the number of moles of AgNO3 and CaCl2 using their respective concentrations and volumes:
Moles of AgNO3 = concentration of AgNO3 × volume of AgNO3
= 0.22 M × 0.1222 L
= 0.026884 moles
Moles of CaCl2 = concentration of CaCl2 × volume of CaCl2
= 0.16 M × 0.1222 L
= 0.019552 moles
Since the molar ratio between AgNO3 and AgCl is 1:1, the number of moles of AgCl formed will be the same as the number of moles of AgNO3.
Therefore, the number of moles of AgCl = 0.026884 moles.
Finally, to calculate the mass of AgCl, we need to use the molar mass of silver chloride (AgCl), which is 143.32 g/mol.
Mass of AgCl = moles of AgCl × molar mass of AgCl
= 0.026884 moles × 143.32 g/mol
= 3.8512 grams
Therefore, approximately 3.85 grams of silver chloride can be prepared by the reaction.