Question

1. Peste 300g solutie Al(OH)3 de c=10% se toarna 200g solutie HCl de c=73%. Se cere:
a) subst in exces si cat este excesul
b) cantitatea de sare formala
La problema de sus trebuie doar punctul b).

2. Un graunte de Al de puritate 80% se introduce in 50g HCl de c=36,5%. Considerand rectia totala,calculati masa granulei de Al

Answer 1

C=md/ms*100=>md=C*ms/100=10*300/100=30g Al(OH)3

C=md/ms*100=>md=73*200/100=146g HCl

MAl(OH)3=78g/mol

MHCl=36,5g/mol

n=m/M=30/78=0,38461 moli Al(OH)3

n=m/M=146/36,5=4 moli HCl

a) Acid in exces -> 4-0,38461=3,615 moli HCl

m acid in exces = 3,615*36,5=131,961g HCl

b) Al(OH)3 + 3HCl -> AlCl3 + 3H2O 

MAlCl3=133.5g/mol

nAl(OH)3=nAlCl3=>mAlCl3(sare)=0,38461*133,5=51,346g AlCl3

2. C=md/ms*100=>md=C*ms/100=36,5*50/100=18,25g HCl

2Al + 6HCl -> 2AlCl3 + 3H2

2*27g Al...............6*36,5g HCl
xg Al.....................18,25g HCl

x=54*18,25/6*36,5=4,5g Al (pur)

puritatea = mpur/m impur*100 => m impur= mpur*100/p = 4,5*100/80=5,625g Al