Answer:
8.415 grams of solid
Step-by-step explanation:
To find the mass of the solid formed in the reaction between calcium chloride (CaCl2) and silver nitrate (AgNO3), you first need to determine the limiting reactant. The limiting reactant is the one that is completely consumed in the reaction, and it determines the maximum amount of product formed.
The balanced chemical equation for the reaction between calcium chloride and silver nitrate is:
CaCl 2 (aq) + 2AgNO3 (aq) --> Ca(NO3)2 (aq) + 2AgCl (s)
From the equation, you can see that one mole of calcium chloride reacts with two moles of silver nitrate to form two moles of silver chloride.
Now, let's calculate the moles of each reactant:
1. Moles of CaCl 2
Moles = M x V
Moles of CaCl2 = 1.230 M x (27.00 x 10^-3)
2. Moles of AgNO3
Moles of AgNO3 = 1.120 M x (35.00 x 10^-3)
Now, determine the limiting reactant by comparing the mole ratios from the balanced equation. The ratio of moles of \( \text{CaCl}_2 \) to \( \text{AgNO}_3 \) is 1:2. Whichever reactant has fewer moles is the limiting reactant.
Calculate the moles formed using the limiting reactant.
Now, you mentioned a typical percent yield of 75%. This means that the actual yield is expected to be 75% of the theoretical yield.
Actual Yield = 0.75 x {Theoretical Yield}
Finally, convert the moles of AgClto grams using the molar mass of ( AgCl) (molar mass of Ag): 107.87 g/mol, molar mass of (Cl): 35.45 g/mol).
Mass = Moles X Molar Mass
Substitute the values and solve for the mass of AgCl formed.
Actual Yield = 0.75 x {Theoretical Yield}
0.06642 mol X (107.87 +2 X 35.45) 8/mol = 11.22
0.75 X 11.22 g = 8.415
Therefore, you would expect to collect approximately 8.415 grams of solid, given a typical per cent yield of 75%.