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

Options

A. 1 mol
B. 2 mol
C. 3 mol
D. No ppt is formed

Answer

B

Explanation

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)`