mdNaOH=500*20/100=100 g n=m/M=100/40=2,5 moli C. adevarat
mH2Odin sol. de NaOH = 500-100 = 400 g
Ca + 2H2O-----> Ca(OH)2 + H2
40g 2*18g 74g
80g y x= 80*74/40=148 gCa(OH)2 A. adevarat
y=80*2*18/40=72
g H2O se consuma deci mai raman 400g-72g=328 g H2O
B.adevarat
msfinal= mdNaOH+mdCa(OH)2+m H2O=100g+ 148g+ 328g=576
cNaOH=100*100/576=7,36 % D. fals
cCa(OH)2=148*100/576=25,69% E. fals