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
millimols AgNO3 = mL x M = 132 mL x 0.719 M = 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 = ?
Post your work if you get stuck.
To find the number of grams of NaCl required to react with the given amount of AgNO3 solution, you need to use the molar ratio between AgNO3 and NaCl.
First, let's calculate the number of moles of AgNO3 in the given solution. We have the volume (132 mL) and the concentration (0.719 M), which means we can use the formula:
moles = concentration × volume
moles of AgNO3 = 0.719 M × 0.132 L (since 132 mL is equal to 0.132 L)
moles of AgNO3 = 0.095208 moles
Next, let's determine the molar ratio between AgNO3 and NaCl from the balanced chemical equation. The ratio is 1:1, meaning one mole of AgNO3 reacts with one mole of NaCl.
Finally, to find the number of grams of NaCl needed, you multiply the number of moles of AgNO3 by the molar mass of NaCl:
grams of NaCl = moles of AgNO3 × molar mass of NaCl
The molar mass of NaCl is 58.44 g/mol.
grams of NaCl = 0.095208 moles × 58.44 g/mol
grams of NaCl = 5.565 grams
Therefore, 5.565 grams of NaCl would be required to react with 132 mL of 0.719 M AgNO3 solution.