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

1 Answer
Dec 8, 2015

Answer:

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#