You mix 55 mL of 1.00 M silver nitrate with 25 mL of 0.81 M sodium chloride. What mass of silver chloride should you form?
To find the mass of silver chloride formed, we need to consider the chemical reaction between silver nitrate (AgNO3) and sodium chloride (NaCl):
AgNO3 + NaCl -> AgCl + NaNO3
From the balanced equation, we can see that 1 mole of silver nitrate reacts with 1 mole of sodium chloride to form 1 mole of silver chloride.
First, let's calculate the number of moles of silver nitrate and sodium chloride used in the reaction.
Number of moles of silver nitrate (AgNO3):
moles = volume (in liters) x molarity
moles = 0.055 L x 1.00 mol/L
moles = 0.055 mol
Number of moles of sodium chloride (NaCl):
moles = volume (in liters) x molarity
moles = 0.025 L x 0.81 mol/L
moles = 0.02025 mol
Since the reaction between silver nitrate and sodium chloride occurs in a 1:1 mole ratio, the limiting reactant is the one that is present in fewer moles. In this case, sodium chloride is the limiting reactant because it has fewer moles (0.02025 mol) compared to silver nitrate (0.055 mol).
To determine the mass of silver chloride formed, we need to calculate the number of moles of silver chloride using the limiting reactant (sodium chloride), and then multiply it by the molar mass of silver chloride.
Molar mass of silver chloride (AgCl):
Ag = 107.87 g/mol
Cl = 35.45 g/mol
AgCl = Ag (1) + Cl (1) = 107.87 g/mol + 35.45 g/mol = 143.32 g/mol
Number of moles of silver chloride:
moles of AgCl = moles of NaCl (because of the 1:1 mole ratio)
moles of AgCl = 0.02025 mol
Mass of silver chloride:
mass = moles x molar mass
mass = 0.02025 mol x 143.32 g/mol
mass = 2.901 g
Therefore, the mass of silver chloride formed is approximately 2.901 grams.
To find the mass of silver chloride formed, we need to determine the limiting reactant first. The limiting reactant is the reactant that is completely consumed and determines the maximum amount of product that can be formed.
Let's calculate the number of moles of silver nitrate and sodium chloride separately:
1. Moles of silver nitrate (AgNO3):
Molarity (M) = moles (mol) / volume (L)
Given:
Molarity of silver nitrate (AgNO3) = 1.00 M
Volume of silver nitrate (AgNO3) = 55 mL = 0.055 L
Using the formula for calculating moles:
Moles of AgNO3 = Molarity × Volume
Moles of AgNO3 = 1.00 mol/L × 0.055 L
Moles of AgNO3 = 0.055 mol
2. Moles of sodium chloride (NaCl):
Molarity (M) = moles (mol) / volume (L)
Given:
Molarity of sodium chloride (NaCl) = 0.81 M
Volume of sodium chloride (NaCl) = 25 mL = 0.025 L
Using the formula for calculating moles:
Moles of NaCl = Molarity × Volume
Moles of NaCl = 0.81 mol/L × 0.025 L
Moles of NaCl = 0.02025 mol
Now, we need to determine the stoichiometry of the reaction between silver nitrate and sodium chloride. The balanced equation for the reaction is:
AgNO3 + NaCl → AgCl + NaNO3
From the balanced equation, we can see that 1 mole of silver nitrate reacts with 1 mole of sodium chloride to form 1 mole of silver chloride.
Since the stoichiometry of the reaction is 1:1, the reactant with the lowest number of moles will be the limiting reactant. In this case, sodium chloride (NaCl) has the lower number of moles.
Next, let's calculate the moles of silver chloride formed using the moles of sodium chloride (NaCl) as the limiting reactant:
Moles of AgCl formed = Moles of NaCl (limiting reactant) × 1
Moles of AgCl formed = 0.02025 mol × 1
Moles of AgCl formed = 0.02025 mol
Finally, we need to convert moles of silver chloride (AgCl) to its mass using the molar mass of silver chloride. The molar mass of AgCl is:
Ag = 107.87 g/mol
Cl = 35.45 g/mol
Molar mass of AgCl = Ag (107.87 g/mol) + Cl (35.45 g/mol)
Molar mass of AgCl = 143.32 g/mol
Mass of AgCl formed = Moles of AgCl × Molar mass of AgCl
Mass of AgCl formed = 0.02025 mol × 143.32 g/mol
Mass of AgCl formed = 2.90 grams
Therefore, the mass of silver chloride formed is 2.90 grams.
This is a limiting reagent problem.
Write and balance the equation.
NaCl + AgNO3 ==> AgCl + NaNO3
moles NaCl = M x L = ?
moles AgNO3 = M x L = ?
Use the coefficients in the balanced equation to convert mol NaCl to mol AgCl.
Do the same for AgNO3.
You get two answers; only one can be right. The correct one in limiting reagent problems is ALWAYS the smaller value and the reagent producing that number is the limiting reagent.
Then convert mol AgCl to grams. g = mols x molar mass.