Question #c882e

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Answer 1 · 2016-10-03T17:07:57

$sf((a))$

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

$sf((b))$

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

$sf((a))$

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 .

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

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