How many grams of CaCl2 would be required to completely react with 310 mL of 0.444 M AgNO3 solution?
Answer in units of grams
Answer in units of grams
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0.310 L * 0.444 mol/L = 0.13764 mol
then it is 2 to1 to CaCl2 so 0.06882 mol
finally
0.06882 mol * 111 g/mol = 7.64 g
then it is 2 to1 to CaCl2 so 0.06882 mol
finally
0.06882 mol * 111 g/mol = 7.64 g