What mass of silver chloride can be prepared by the reaction of 160.0 mL of 0.10 M silver nitrate with 120.0 mL of 0.19 M calcium chloride?
To find the mass of silver chloride that can be prepared by the reaction, you first need to determine the limiting reagent. The limiting reagent is the reactant that is completely consumed first, and it determines the maximum amount of product that can be formed.
In this case, you have silver nitrate and calcium chloride reacting to form silver chloride.
First, calculate the moles of silver nitrate:
Moles of silver nitrate = volume (in L) × concentration
Moles of silver nitrate = 0.160 L × 0.10 mol/L
Moles of silver nitrate = 0.016 mol
Next, calculate the moles of calcium chloride:
Moles of calcium chloride = volume (in L) × concentration
Moles of calcium chloride = 0.120 L × 0.19 mol/L
Moles of calcium chloride = 0.0228 mol
The balanced chemical equation for the reaction is:
AgNO3 + CaCl2 → AgCl + Ca(NO3)2
From the balanced equation, the stoichiometry of the reaction is 1:2. This means that for every 1 mole of silver nitrate, 2 moles of silver chloride are formed.
To determine the limiting reagent, compare the molar ratios of the two reactants. The mole ratio is:
Moles of silver nitrate : Moles of calcium chloride
0.016 mol : 0.0228 mol
Since the molar ratio is less than 1:2, it means that silver nitrate is the limiting reagent.
To find the moles of silver chloride formed, use the mole ratio from the balanced equation:
Moles of silver chloride = Moles of silver nitrate × (2 moles of silver chloride / 1 mole of silver nitrate)
Moles of silver chloride = 0.016 mol × (2 mol/1 mol)
Moles of silver chloride = 0.032 mol
Finally, calculate the mass of silver chloride:
Mass of silver chloride = Moles of silver chloride × molar mass of silver chloride
Mass of silver chloride = 0.032 mol × (107.87 g/mol)
Mass of silver chloride = 3.45 g
Therefore, the mass of silver chloride that can be prepared is 3.45 grams.
To determine the mass of silver chloride that can be prepared, we first need to write and balance the chemical equation for the reaction between silver nitrate (AgNO3) and calcium chloride (CaCl2).
The chemical equation is: AgNO3 + CaCl2 -> AgCl + Ca(NO3)2
From the equation, we can see that the reaction will produce silver chloride (AgCl).
To solve the problem, we will use the concept of stoichiometry, which relates the amount of reactants and products in a chemical reaction.
Step 1: Calculate the moles of silver nitrate (AgNO3) and calcium chloride (CaCl2) using the given concentrations and volumes.
Moles of AgNO3 = concentration of AgNO3 × volume of AgNO3
= 0.10 M × 0.160 L
= 0.016 mol
Moles of CaCl2 = concentration of CaCl2 × volume of CaCl2
= 0.19 M × 0.120 L
= 0.0228 mol
Step 2: Determine the stoichiometric ratio between the reactants and product based on the balanced chemical equation.
From the balanced chemical equation, we can see that 1 mole of AgNO3 reacts with 2 moles of AgCl.
Step 3: Calculate the limiting reactant (the reactant that is completely consumed first) by comparing the moles of the reactants with the stoichiometric ratio.
Since the stoichiometric ratio between AgNO3 and AgCl is 1:2, we can conclude that 0.016 mol of AgNO3 will produce 0.032 mol of AgCl.
Similarly, since the stoichiometric ratio between CaCl2 and AgCl is 1:1, 0.0228 mol of CaCl2 will produce 0.0228 mol of AgCl.
From the calculations, we can see that CaCl2 is the limiting reactant since it produces fewer moles of AgCl (0.0228 mol) compared to AgNO3 (0.032 mol).
Step 4: Calculate the mass of AgCl produced using the moles of AgCl and its molar mass.
The molar mass of AgCl is the sum of the atomic masses of silver (Ag) and chlorine (Cl), which is (107.87 g/mol + 35.45 g/mol = 143.32 g/mol).
Mass of AgCl = moles of AgCl × molar mass of AgCl
= 0.0228 mol × 143.32 g/mol
≈ 3.27 g
Therefore, the mass of silver chloride that can be prepared by the reaction of 160.0 mL of 0.10 M silver nitrate with 120.0 mL of 0.19 M calcium chloride is approximately 3.27 grams.