24.8 mL sample of a 4.63 x 10–2 M solution of NaCl is mixed with a 30.0 mL sample of a 3.67 x 10–2 M solution of AgNO3. What is the mass, in grams, of AgCl precipitate that forms?
write the chemical reaction, balance it.
determine the moles of each reactant present, then find the limiting reactant.
from the limiting reactant, use the mole ratio to determine moles of AgCl,convert to mass in grams.
1.15 g
To find the mass of AgCl precipitate that forms, we need to understand the chemical reaction that occurs when NaCl and AgNO3 are mixed. In this case, NaCl reacts with AgNO3 to form AgCl precipitate.
The balanced chemical equation for this reaction is:
NaCl + AgNO3 → AgCl + NaNO3
From the equation, we can see that 1 mole of NaCl reacts with 1 mole of AgNO3 to produce 1 mole of AgCl.
To find the moles of NaCl and AgNO3 in the given solutions, we'll use the formula:
moles = concentration (M) x volume (L)
For the NaCl solution:
Moles of NaCl = concentration of NaCl x volume of NaCl solution
= (4.63 x 10^(-2) M) x (24.8 mL / 1000 mL/L)
= 0.01144 moles
For the AgNO3 solution:
Moles of AgNO3 = concentration of AgNO3 x volume of AgNO3 solution
= (3.67 x 10^(-2) M) x (30.0 mL / 1000 mL/L)
= 0.011 moles
Since the reaction between NaCl and AgNO3 is 1:1, the moles of AgCl formed will be equal to the limiting reactant, which is NaCl in this case. Therefore, the moles of AgCl formed will also be 0.011 moles.
To find the mass of AgCl precipitate formed, we'll use its molar mass, which is 143.32 g/mol. Using the formula:
mass = moles x molar mass
Mass of AgCl = 0.011 moles x 143.32 g/mol
= 1.576 g
Therefore, the mass of AgCl precipitate that forms is 1.576 grams.