n H2=\(\dfrac{8,96}{22,4}\)=0,4 mol
2Al + 6HCl→2AlCl3 + 3H2 (1)
x----------------------------3\2x
Zn +2 HCl→ZnCl2 + H2 (2)
y----------------------------y mol
ta có hệ :
27x+65y=11,9
\(\dfrac{3}{2}\)x+y=0,4
=>x=0,2 mol->m Al=0,2.27=5,4g
=>y=0,1 mol->m Zn=0,1.65=6,5g
nắm chắc pthh là xong hết , về luyện là đc
\(nH_2=8,96:22,4=0,4mol\)
PTHH:
\(2Al+2Zn+2HCl\rightarrow2AlZnCl+H_2\)
0,8<--0,8<----0,8<------<0,8--------<0,4
\(mAl=0,8.27=21,6gam\)
\(mZn=0,8.65=52gam\)
\(mAlZnCl=102gam\)
PTHH : 2Al + 6HCl -> 2AlHCl3 + 3H2
PTHH : Zn + HCl -> ZnCl2 + H2
\(n_{H_2}=\dfrac{8.96}{22.4}=0,4\left(mol\right)\)
Gọi Al là a ; Zn là b
Ta có
\(\left\{{}\begin{matrix}27a+65b=11,9\\a+b=0,4\end{matrix}\right.\)
\(=>a=0,2\left(mol\right)->m_{Al}=0,2.27=5,4\left(g\right)\)
\(=>b=0,1\left(mol\right)->m_{Zn}=0,1.65=6,5\left(g\right)\)
PTHH : 2Al + 6HCl -> 2AlHCl3 + 3H2
PTHH : Zn + HCl -> ZnCl2 + H2
\(n_{H_2}=\dfrac{8.96}{22.4}=0,4\left(mol\right)\)
Gọi Al là a ; Zn là b
Ta có
\(\left\{{}\begin{matrix}27a+65b=11,9\\a+b=0,4\end{matrix}\right.\)
\(=>a=0,2\left(mol\right)->m_{Al}=0,2.27=5,4\left(g\right)\)
\(=>b=11,9-5,4=6,5\left(g\right)\)