Question #c882e

1 Answer
Oct 3, 2016

Answer:

#sf((a))#

#sf([HCl]=0.5056color(white)(x)"mol/l")#

#sf((b))#

#sf([RbOH]=0.9118color(white)(x)"mol/l")#

Explanation:

#sf((a))#

Start with the equation:

#sf(Ba(OH)_(2)+2HClrarrBaCl_2+2H_2O)#

#sf(c=n/v)#

#:.##sf(n=cxxv)#

#:.##sf(n_(Ba(OH)_2)=0.1529xx43.09=6.5884color(white)(x)"mmol")#

From the equation we can see that the no. moles of #sf(HCl)# must be twice this.

#:.##sf(n_(HCl)=6.5884xx2=13.177color(white)(x)"mmol")#

#:.##sf([HCl]=c/v=(13.177xxcancel(10^(-3)))/(26.06/(cancel(1000)))=0.5056color(white)(x)"mol/l")#

#sf((b))#

#sf(HCl+RbOHrarrRbCl+H_2O)#

#sf(n_(HCl)=cxxv=0.5056xx27.05=13.676color(white)(x)"mmol")#

From the equation we can see that the no. moles of #sf(RbOH)# must be the same .

#:.##sf(n_(RbOH)=13.676color(white)(x)"mmol")#

#:.##sf([RbOH]=c/v=(13.676xxcancel(10^(-3)))/(15.00/(cancel(1000)))=0.9118color(white)(x)"mol/l")#