1.

How many moles of AgCl would be obtained, when Co(NH_(3) )_(5)Cl_(3) is treated with excess of AgNO_(3)?

Answer»

1 mol
2 mol
3 mol
No ppt is formed

Solution :No. of MOLE.s `=0.1 XX 0.1 = 0.01`
`[Co(NH_(3))_(5)Cl]Cl_(2) to [Co(NH_(3))_(5)Cl]^(+2) +2CL^(-)`
`underset(("EXCESS"))overset(0.01" mole")(Ag^(+))underset((0.02" mole"))overset(0.01" mole")(+Cl^(-)""to) underset((0.02" mole")) overset(0.02" mole" to)(AGCL)`


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