Question

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5.4 g pulbere de aluminiu reactioneaza cu HCL . numarul de moli de HCL necesari , masa de sare rezultata si volumul de gaz rezultat au valorile :

Answer 1

2Al + 6HCl ⇒ 2AlCl₃ + 3H₂↑

MAl=27g/mol

MHCl=AH+ACl=1+35,5=36,5g/mol

MAlCl₃=AAl+3ACl=27+3*35,5=133.5g/mol

n=m/M=5,4/27=0,2 moli Al

2 mol Al..................6 moli HCl
0,2 moli Al.............x moli HCl

x=0,2*6/2=Calculeaza !!

2 moli Al.......................2*133,5g AlCl₃
0,2 moli Al............................xg AlCl₃

x=2*133,5*0,2/2 = Calculeaza !!

2 moli Al..........................3*22,4L H₂
0,2 moli Al............................xL H₂

x=67,2*0,2/2=6,72 L H₂