Question

300g Al de puritate 90% reactioneaza cu o solutie de HCl 36,5%.
a) m solutiei de HCl
b) nr. de moli de gaz degajat

Answer 1

2Al+6HCl = 3H2 + 2AlCl
P supra 100 = m pur supra m impur => m pur = (90*300):100 = 270 g
M Al = 27 u.a.m => u = 27 g/mol
M HCl = A H +A Cl = 36.5 u.a.m => u = 36.5 g/ mol
54:270 =219:x => x = 270*219:54 = 1095g HCl
C:100=md:ms=>ms=100*md:C <=> 100*1095:36,5=3000
MH2 =2AH=2 u.a.m=>u=2g/mol
54:270=6*y=>y=270*6:54=30g
nr.moli=Vgaz:Vmolar=>m:u=Vgaz:Vmolar=Vgaz=m*Vmolar:u=30*22.4:2=336 m3