ta co pthh
2Al + 6HCl \(\rightarrow\) 2AlCl3 + 3H2
Theo de bai ta co
nAl=\(\dfrac{5,4}{27}=0,2mol\)
a,Theo pthh
nH2=\(\dfrac{3}{2}nAl=\dfrac{3}{2}.0,2=0,3mol\)
\(\Rightarrow VH2_{\left(dktc\right)}=0,3.22,4=6,72l\)
b,Theo pthh
nHCl=\(\dfrac{6}{2}nAl=\dfrac{6}{2}.0,2=0,6mol\)
\(\Rightarrow mct=mHCl=0,6.36,5=21,9g\)
\(\Rightarrow\) Nong do %cua dd HCl la
C%=\(\dfrac{mct}{mdd}.100\%=\dfrac{21,9}{200}.100\%=10,95\%\)
c,Cach 1
Theo pthh
nAlCl3=nAl=0,2 mol
\(\Rightarrow mAlCl3=0,2.133,5=26,7g\)
Cach 2
Ap dung DLBTKL ta co
mAl + mHCl=mAlCl3+mH2
\(\Rightarrow\) mAlCl3=(mAl+mHCl)-mH2=(5,4+21,9)-(0,3.2)=26,7g
nAl=m/M=5,4/27=0,2(mol)
PT:
2Al + 6HCl -> 2AlCl3 + 3H2\(\uparrow\)
2..........6...........2...............3 (mol)
0,2 ->0,6 ->0,2 -> 0,3 (mol)
VH2=n.22,4=0,3.22,4=6,72(lít)
b) mHCl=n.M=0,6.36,5=21,9 (g)
=> \(C\%_{HCl}=\dfrac{m_{HCl}.100\%}{m_{ddHCl}}=\dfrac{21,9.100}{200}=10,95\left(\%\right)\)
c) Cách 1: Theo PTHH:
mAlCl3=n.M=0,2. 133,5=26,7(g)
Cách 2: Áp dụng ĐLBTKL:
mAl + mHCl = mAlCl3 + mH2
=> mAlCl3=mAl + mHCl - mH2= 5,4+21,9 - (0,3.2)=26,7 ( g)
Nhớ tick nha , chúc bạn học tốt
nAl = 5.4/27 =0.2(mol)
Ta có pthh: 2Al +6HCl--->2AlCl3 + 3H2
(mol) 0.2 0.6 0.2 0.3
=>VH2 = 0.3 *22.4=6.72(l)
=>mHCl = 0.6*36.5=21.9(g)
C%dd HCl = 21.9/200+21.9 * 100% = 9.87%(xấp xỉ nha)
mAlCl3 = 5.4+21.9-(0.3*2)=26.7 (g)