- #1
Lori
So my answer is off by about 1.16 grams and am wondering if i did something wrong?
Question: What mass of silver iodide precipitates when 25.0 ml of 2.30 M silver acetate solution is mixed with 10.0 ml of 2.35 M calcium iodide solution?
Given:
0.025 L of 2.30 M AgC2H3O2 = 0.0575 mols acetate
0.01 L of 2.35 M caI2 = 0.0235 mols Calcium iodide
Want:
? grams of AgI2
So, i converted 0.0575 mols of acetate to silver and got 6.20 grams Ag
by 0.0575 mols acetate * (1mol Ag/1mol Acetate) *107.87 grams/mol Ag = 6.20 grams Ag
For 0.0235 mols CaI2, i got 5.96 I grams by doing (0.0235 grams CaI2) * (2 mols I/ 1 mol Ca) *126.9 gram/mol...
i added 5.96 grams of Iodine and 6.20 grams of silver together and got 12.17 grams AgI2?
Did i miss something
Question: What mass of silver iodide precipitates when 25.0 ml of 2.30 M silver acetate solution is mixed with 10.0 ml of 2.35 M calcium iodide solution?
Given:
0.025 L of 2.30 M AgC2H3O2 = 0.0575 mols acetate
0.01 L of 2.35 M caI2 = 0.0235 mols Calcium iodide
Want:
? grams of AgI2
So, i converted 0.0575 mols of acetate to silver and got 6.20 grams Ag
by 0.0575 mols acetate * (1mol Ag/1mol Acetate) *107.87 grams/mol Ag = 6.20 grams Ag
For 0.0235 mols CaI2, i got 5.96 I grams by doing (0.0235 grams CaI2) * (2 mols I/ 1 mol Ca) *126.9 gram/mol...
i added 5.96 grams of Iodine and 6.20 grams of silver together and got 12.17 grams AgI2?
Did i miss something