What weight of AgCl can be obtained by precipitating all of the ...(cntd)?
2015-10-20 4:03 pm
...Of the Ag+ from 50mL of 0.12M AgNO3 when added in NaCl solution? (Answer is 0.86g)
回答 (1)
2015-10-20 4:18 pm
Equation:
Ag+(aq) + Cl-(aq) → AgCl(s)
1mol Ag+ will produce 12 mol AgCl.
Mol Ag+ in 50mL of 0.12M AgCl solution = 50/1000*0.12 = 0.006 mol Ag+
The reaction will produce 0.006 mol AgCl
Molar mass AgCl = 107.87 + 35.45 = 143.32g/mol
Mass of 0.006 mol = 0.006*143.32 = 0.86g.
Ag+(aq) + Cl-(aq) → AgCl(s)
1mol Ag+ will produce 12 mol AgCl.
Mol Ag+ in 50mL of 0.12M AgCl solution = 50/1000*0.12 = 0.006 mol Ag+
The reaction will produce 0.006 mol AgCl
Molar mass AgCl = 107.87 + 35.45 = 143.32g/mol
Mass of 0.006 mol = 0.006*143.32 = 0.86g.
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