What volume of 0.123 M AgNO3(aq) is needed to form 0.657 g of AgCl(s)?
0.564 mL.
18.6 mL.
37.2 mL.
1.77 L.
To solve this question, we need to use the concept of stoichiometry and the balanced chemical equation for the reaction between silver nitrate (AgNO3) and silver chloride (AgCl).
The balanced chemical equation for the reaction is:
AgNO3(aq) + NaCl(aq) -> AgCl(s) + NaNO3(aq)
From the equation, we can see that one mole of AgNO3 reacts with one mole of AgCl.
First, we need to calculate the number of moles of AgCl formed. The molar mass of AgCl is the sum of the atomic masses of silver (Ag) and chlorine (Cl), which is 107.87 g/mol.
Number of moles of AgCl = mass of AgCl / molar mass of AgCl
Number of moles of AgCl = 0.657 g / 107.87 g/mol
Number of moles of AgCl = 0.00608 mol
Since the stoichiometry of the reaction is 1:1, the number of moles of AgNO3 needed is also 0.00608 mol.
Next, we can use the formula for the concentration of a solution:
Concentration = moles / volume
We can rearrange this formula to solve for the volume:
Volume = moles / concentration
Now, we need to use the molarity of AgNO3 (0.123 M) to find the volume.
Volume of AgNO3 = 0.00608 mol / 0.123 mol/L
Volume of AgNO3 = 0.0494 L
Finally, we convert the volume from liters to milliliters:
Volume of AgNO3 = 0.0494 L x 1000 mL/L
Volume of AgNO3 = 49.4 mL
Therefore, the volume of 0.123 M AgNO3(aq) needed to form 0.657 g of AgCl(s) is 49.4 mL.
To determine the volume of 0.123 M AgNO3(aq) needed, we can use the stoichiometry of the balanced chemical equation for the reaction:
AgNO3(aq) + NaCl(aq) -> AgCl(s) + NaNO3(aq)
From the balanced equation, we can see that 1 mole of AgNO3 reacts with 1 mole of AgCl.
First, we need to calculate the number of moles of AgCl using its molar mass:
molar mass of AgCl = 107.87 g/mol
moles of AgCl = mass / molar mass = 0.657 g / 107.87 g/mol ≈ 0.006076 mol
Since the stoichiometry is 1:1, we know that we need 0.006076 moles of AgNO3 for the reaction.
Now, we can calculate the volume of 0.123 M AgNO3 solution using the formula:
volume = moles / concentration
volume = 0.006076 mol / 0.123 mol/L ≈ 0.04938 L
Converting from liters to milliliters:
volume = 0.04938 L * 1000 mL/L = 49.38 mL
Therefore, the volume of 0.123 M AgNO3(aq) needed is approximately 49.38 mL. None of the options provided match this answer, so it seems there might be an error in the given choices.
Write the equation.
AgNO3 ==> AgCl (obviously 1 mol AgNO3 is need to form 1 mole AgCl).
0.657 g AgCl is how many moles? moles = g/molar mass = ??
moles AgCl = moles AgNO3.
Molarity = moles/L.
You know molarity and moles, solve for L.