What mass of precipitate is formed when 0.750 L of 2.50 mol/L CaCl2 reacts with 650 mL of 1.75 mol/L AgNO3
You have a limiting reagent problem here. You know that because amounts are give for BOTH of the reactants.
2AgNO3 + CaCl2 ==> 2AgCl(s) + Ca(NO3)2
mols AgNO3 = M x L = ?
mols CaCl2 = M x L = ?
Using the coefficients in the balanced equation, convert mols AgNO3 to mols AgCl.
Do the same and convert mols CaCl2 to mols AgCl.
It is likely that the two values will not agree which means one of them is wrong; the correct value in LR problems is ALWAYS the smaller one and the reagent responsible for that is the LR.
Using the smaller value for mols AgCl, convert to grams. g AgCl = mols AgCl x molar mass AgCl.
How do i convert mols cacl2 to agcl
You use the coefficients in the balanced equation to convert anything you wish. For CaCl2 to mols AgCl it is
?mols CaCl2 x (2 mols AgCl/1 mol CaCl2) = ?mols CaCl2 x 2/1 = ?mols AgCl.
I don't understand...
How did you convert mols AgNO3 to mols AgCl?
To determine the mass of precipitate formed, we need to first identify the balanced chemical equation for the reaction between calcium chloride (CaCl2) and silver nitrate (AgNO3).
The balanced chemical equation for the reaction is:
CaCl2 + 2AgNO3 -> 2AgCl + Ca(NO3)2
From the balanced equation, we can see that 1 mole of CaCl2 reacts with 2 moles of AgNO3 to produce 2 moles of AgCl.
Step 1: Calculate the number of moles of CaCl2 and AgNO3 used.
The volume of the CaCl2 solution is given as 0.750 L with a concentration of 2.50 mol/L. To find the number of moles of CaCl2 used, we multiply the volume by the concentration:
moles of CaCl2 = volume of CaCl2 solution * concentration of CaCl2
= 0.750 L * 2.50 mol/L
= 1.875 mol
The volume of the AgNO3 solution is given as 650 mL, which is equivalent to 0.650 L, with a concentration of 1.75 mol/L. To find the number of moles of AgNO3 used, we multiply the volume by the concentration:
moles of AgNO3 = volume of AgNO3 solution * concentration of AgNO3
= 0.650 L * 1.75 mol/L
= 1.1375 mol
Step 2: Determine the limiting reagent.
To identify the limiting reagent, we compare the mole ratios of CaCl2 and AgNO3 from the balanced equation. According to the reaction stoichiometry, 1 mole of CaCl2 reacts with 2 moles of AgNO3 to form 2 moles of AgCl.
The moles ratio of CaCl2 to AgNO3 is 1:2. Since we have 1.875 moles of CaCl2 and 1.1375 moles of AgNO3, we need to consider the ratio 1.875:1.1375.
Comparing the ratios, it is clear that AgNO3 is in excess. Hence, CaCl2 is the limiting reagent.
Step 3: Calculate the moles of AgCl formed.
Since CaCl2 is the limiting reagent, we can use its moles to calculate the moles of AgCl formed in the reaction.
From the balanced equation, we know that 1 mole of CaCl2 reacts to produce 2 moles of AgCl.
Therefore, 1.875 moles of CaCl2 will produce 2 * 1.875 = 3.75 moles of AgCl.
Step 4: Calculate the mass of AgCl formed.
The molar mass of AgCl is the sum of the atomic masses of silver (Ag) and chlorine (Cl):
molar mass of AgCl = atomic mass of Ag + atomic mass of Cl
= 107.87 g/mol + 35.45 g/mol
= 143.32 g/mol
To calculate the mass of AgCl formed, we multiply the moles of AgCl by its molar mass:
mass of AgCl = moles of AgCl * molar mass of AgCl
= 3.75 mol * 143.32 g/mol
= 536.7 g
Therefore, the mass of precipitate (AgCl) formed in the reaction is 536.7 grams.