How many milliliters of 0.277 M AgNO3 solution are needed to react completely with 47.1 mL of 0.278 M NaCl solution? How many grams of AgCl will be formed? The reaction is
NaCl(aq) + AgNO3(aq) → AgCl(s) + NaNO3(aq)
Use the procedure in the succeeding problem [HNO3 and Ba(OH)2] to solve this one. Also you may need the stoichiometry procedure too. Here is the link again.
http://www.jiskha.com/science/chemistry/stoichiometry.html
To find out how many milliliters of AgNO3 solution are needed to react with the NaCl solution, we can use the stoichiometry of the balanced chemical equation.
First, let's write the balanced chemical equation:
NaCl(aq) + AgNO3(aq) → AgCl(s) + NaNO3(aq)
From the equation, we can see that the mole ratio between NaCl and AgNO3 is 1:1. This means that 1 mole of NaCl reacts with 1 mole of AgNO3.
To find the moles of NaCl, we can use the concentration and volume of the NaCl solution:
moles of NaCl = concentration × volume
moles of NaCl = 0.278 M × 47.1 mL
moles of NaCl = 0.01306 moles
Since the mole ratio between NaCl and AgNO3 is 1:1, we need the same number of moles of AgNO3 to react:
moles of AgNO3 = 0.01306 moles
Now, let's find the volume of the AgNO3 solution in milliliters needed to have 0.01306 moles of AgNO3.
volume of AgNO3 = moles of AgNO3 / concentration of AgNO3
volume of AgNO3 = 0.01306 moles / 0.277 M
volume of AgNO3 = 0.0471 L
Converting from liters to milliliters:
volume of AgNO3 = 0.0471 L × 1000 mL/L
volume of AgNO3 = 47.1 mL
Therefore, 47.1 mL of 0.277 M AgNO3 solution is needed to react completely with 47.1 mL of 0.278 M NaCl solution.
Now, let's calculate the mass of AgCl formed.
From the stoichiometry of the balanced chemical equation, we can see that the mole ratio between AgNO3 and AgCl is 1:1. This means that 1 mole of AgNO3 produces 1 mole of AgCl.
moles of AgCl = moles of AgNO3
moles of AgCl = 0.01306 moles
To find the mass of AgCl, we can use the molar mass of AgCl:
mass of AgCl = moles of AgCl × molar mass of AgCl
mass of AgCl = 0.01306 moles × (107.87 g/mol)
mass of AgCl = 1.39 grams
Therefore, 1.39 grams of AgCl will be formed.
To find the number of milliliters of 0.277 M AgNO3 solution needed to react completely with 47.1 mL of 0.278 M NaCl solution, we can use the stoichiometry and the concept of balanced equations.
Step 1: Write the balanced equation for the reaction:
NaCl(aq) + AgNO3(aq) → AgCl(s) + NaNO3(aq)
Step 2: Determine the mole ratio of AgNO3 to NaCl from the balanced equation. In this case, it is 1:1.
Step 3: Calculate the number of moles of NaCl:
moles of NaCl = volume of NaCl solution (in liters) × concentration of NaCl (in moles per liter)
moles of NaCl = 47.1 mL ÷ 1000 mL/L × 0.278 moles/L
Step 4: Based on the stoichiometry, the number of moles of AgNO3 required will be the same as the moles of NaCl.
Step 5: Calculate the volume of AgNO3 solution required:
volume of AgNO3 solution (in liters) = moles of AgNO3 ÷ concentration of AgNO3 (in moles per liter)
volume of AgNO3 solution = moles of NaCl ÷ 0.277 moles/L
Step 6: Convert the volume in liters to milliliters:
volume of AgNO3 solution = volume of AgNO3 solution × 1000 mL/L
Step 7: Calculate the grams of AgCl formed:
moles of AgCl = moles of NaCl (since they have a 1:1 stoichiometric ratio)
mass of AgCl = moles of AgCl × molar mass of AgCl
Now you can plug in the values and calculate the results.