se afla md sol. de HCl
md= c . ms : 100
md= 36,5 . 400 : 100= 146 g HCl
se scriu pe ec. datele problemei
xg 146g yg
NaOH + HCl = NaCl + H2O
40g 36,5 g 58,5 g
M HCl=1+ 35,5=36,5-----> 1mol=36,5g
MNaOH=23+16+1=40----->1mol=40g
MNaCl=23+35,5=58,5----->1mol=58,5g
x=40.146 : 36,5=160 g NaOH [ 160g: 40g/moli=4moli NaOH ]
y= 146.58,5 : 36,5=234 g NaCl [ 234g : 58,5g/moli=4moli NaCl ]