What weight of agcl will be precipitated when a solution containing 4.77g of nacl is?
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Equation:
Ag+ + Cl- = AgCl
1 mole of Ag+ reacts with 1 mole of NaCl
Molar mass of NaCl = 23 + 35.5 = 58.5 g/mol
molar mass AgCl = 108 + 35.5 = 143.5 g/mol
If 58.5 g NaCl precipitate 143.5 g AgCl
∴ 4.77 g = (4.77x143.5)/58.5
= 11.7 g
Ag+ + Cl- = AgCl
1 mole of Ag+ reacts with 1 mole of NaCl
Molar mass of NaCl = 23 + 35.5 = 58.5 g/mol
molar mass AgCl = 108 + 35.5 = 143.5 g/mol
If 58.5 g NaCl precipitate 143.5 g AgCl
∴ 4.77 g = (4.77x143.5)/58.5
= 11.7 g
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