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.
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.
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