a)
ms NaOH=md+mH2O=4g+46g=50g
c=md*100/ms=4*100/50=8 %
b)
NaOH+HCl--->NaCl+H2O
mdHCl=ms*c/100=50*36,5/100=0,5*36,5g HCl=18,25 g HCl
din ecuatia reactiei;
40g NaOH.......36,5gHCl........58,5g NaCl
4g NaOH..........x gHCl.........y g NaCl
x=3,65g HCl se consuma
deci HCl este in exces cu 18,25g-3,65g=14,6 g
y=4*58,5/40=5,85 g NaCl