Silver nitrate (AgNO3) reacts with sodium
chloride as indicated by the equation
AgNO3 + NaCl → AgCl + NaNO3 .
How many grams of NaCl would be required
to react with 132 mL of 0.719 M AgNO3
solution?
Answer in units of grams.
AgNO3 + NaCl → AgCl + NaNO3
millimoles AgNO3 = 132 mL x 0.719 = 94.9
millimols NaCl needed = 94.9 x (1 mol NaCl/1 mol AgNO3) = 94.9
mols NaCl = 0.0949.
grams NaCl = mols NaCl x molar mass NaCl = ?
To find the grams of NaCl required to react with 132 mL of 0.719 M AgNO3 solution, we can use the balanced chemical equation and stoichiometry.
First, let's determine the moles of AgNO3 in the given volume of solution:
Molarity (M) = moles of solute / liters of solution
0.719 M = moles of AgNO3 / 0.132 L
moles of AgNO3 = 0.719 M * 0.132 L = 0.094728 moles
From the balanced chemical equation, we can see that 1 mole of AgNO3 reacts with 1 mole of NaCl. Therefore, the moles of NaCl required will also be 0.094728 moles.
Now, we need to convert moles of NaCl to grams using its molar mass:
Molar mass of NaCl = 22.99 g/mol (sodium) + 35.45 g/mol (chlorine)
= 58.44 g/mol
grams of NaCl = moles of NaCl * molar mass of NaCl
= 0.094728 moles * 58.44 g/mol
= 5.53 grams
Therefore, 5.53 grams of NaCl would be required to react with 132 mL of 0.719 M AgNO3 solution.
To determine the number of grams of NaCl required to react with 132 mL of 0.719 M AgNO3 solution, we can use the following steps:
Step 1: Calculate the number of moles of AgNO3 present in the given volume of solution.
First, we need to convert the volume of the solution from milliliters (mL) to liters (L):
132 mL = 132/1000 = 0.132 L
Next, we can calculate the number of moles of AgNO3 using the molarity (M) and volume (V) of the solution:
Number of moles of AgNO3 = Molarity × Volume
= 0.719 mol/L × 0.132 L
= 0.094908 mol
Step 2: Use the balanced chemical equation to determine the stoichiometry between AgNO3 and NaCl.
According to the balanced chemical equation:
AgNO3 + NaCl → AgCl + NaNO3
From the equation, we can see that 1 mole of AgNO3 reacts with 1 mole of NaCl.
Step 3: Calculate the number of moles of NaCl required.
Since the stoichiometry between AgNO3 and NaCl is 1:1, the number of moles of NaCl required is also 0.094908 mol.
Step 4: Convert moles of NaCl to grams.
To convert moles of NaCl to grams, we need to know the molar mass of NaCl.
Molar mass of NaCl = atomic mass of Na + atomic mass of Cl
= (22.99 g/mol) + (35.45 g/mol)
= 58.44 g/mol
Finally, we can calculate the number of grams of NaCl using the number of moles and the molar mass:
Number of grams of NaCl = Number of moles × Molar mass
= 0.094908 mol × 58.44 g/mol
≈ 5.53 grams
Therefore, approximately 5.53 grams of NaCl would be required to react with 132 mL of 0.719 M AgNO3 solution.