Question

How many milliliters of 0.200M `NaOH` solution are needed to react with 22.0 mL of a 0.490 M `NiCl_2` solution?

Answer 1

There are `0.0108` `mol` `NiCl_2`. `0.0216` `mol` `NaOH` are required, given `0.200` `mol*L^-1` `NaOH(aq)`.

Explanation:

`NiCl_2(aq) + 2NaOH(aq) rarr Ni(OH)_2(s)darr + 2NaCl(aq)`

There are `0.0108` `mol` `NiCl_2` (`22.0xx10^(-3)cancelLxx0.490*mol*cancel(L^-1)`). So, by stoichiometry `0.0216` `mol` `NaOH` are required.

There is `0.200` `mol*L^-1` `NaOH` solution available, so...

`(0.0216*cancel(mol))/(0.200*cancel(mol)*cancel(L^-1))xx1000*mL*cancel(L^-1)` `=` `?? mL`