xg 146 g yg zg
2Al + 6HCl = 2AlCl3 + 3 H2
2.27g 6.36,5g 2.133,5g 3.2g
MHCl=1+35,5=36,5 -------->1mol=36,5 g
MAlCl3=27+3 .35,5=133,5-------->1mol=133,5 g
se afla md sol. de HCl
c=md.100: ms
md=c.ms :100
md=730.20:100=146g HCl
se calculeaza x
x=146.2 .27 :6.36,5=36 g Al
se afla y din ec.
y=146.2 .133,5 :6 .36,5=178 g AlCl3
se afla z din ec.
z=146. 3.2 :6.36,5=4 g hidrogen