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\(n_{Zn}=\dfrac{6.5}{65}=0.1\left(mol\right)\)
\(Zn+2HCl\rightarrow ZnCl_2+H_2\)
\(0.1.................................0.1\)
\(Đặt:n_{CuO\left(pư\right)}=x\left(mol\right)\)
\(CuO+H_2\underrightarrow{t^0}Cu+H_2O\)
\(x............x\)
\(m_{cr}=6-80x+64x=5.2\left(g\right)\)
\(\Rightarrow x=0.05\)
\(H\%=\dfrac{0.05}{0.075}\cdot100\%=66.67\%\)
\(n_{HCl}=0.5\cdot2=1\left(mol\right)\)
\(2Al+6HCl\rightarrow2AlCl_3+3H_2\)
\(..........1.........\dfrac{1}{3}.......0.5\)
\(V_{H_2}=0.5\cdot22.4=11.2\left(l\right)\)
\(m_{AlCl_3}=\dfrac{1}{3}\cdot133.5=44.5\left(g\right)\)
\(CuO+H_2\underrightarrow{^{t^0}}Cu+H_2O\)
\(0.5.....0.5\)
\(m_{CuO}=0.5\cdot80=40\left(g\right)\)
Zn + 2HCl -> ZnCl2 + H2 (1)
nZn=0,1(mol)
Từ 1:
nZnCl2=nH2=nZn=0,1(mol)
mZnCl2=136.0,1=13,6(g)
VH2=0,1.22,4=2,24(lít)
CuO +H2 -> Cu + H2O (2)
Từ 2:
nO=nH2=0,1(mol)
mO=16.0,1=1,6(g)
mchất rắn còn lại=10-1,6=8,4(g)
Chúc Bạn Học Tốt
a, \(n_{Zn}=\dfrac{13}{65}=0,2\left(mol\right)\)
\(n_{HCl}=\dfrac{18,25}{36,5}=0,5\left(mol\right)\)
PT: \(Fe+2HCl\rightarrow FeCl_2+H_2\)
Xét tỉ lệ: \(\dfrac{0,2}{1}< \dfrac{0,5}{2}\), ta được HCl dư.
Theo PT: \(n_{H_2}=n_{Fe}=0,2\left(mol\right)\Rightarrow V_{H_2}=0,2.22,4=4,48\left(l\right)\)
b, - Nhúng quỳ tím vào dd thấy quỳ chuyển đỏ do HCl dư.
c, \(n_{CuO}=\dfrac{24}{80}=0,3\left(mol\right)\)
PT: \(H_2+CuO\underrightarrow{t^o}Cu+H_2O\)
Xét tỉ lệ: \(\dfrac{0,2}{1}< \dfrac{0,3}{1}\), ta được CuO dư.
Theo PT: \(n_{CuO\left(pư\right)}=n_{Cu}=n_{H_2}=0,2\left(mol\right)\Rightarrow n_{CuO\left(dư\right)}=0,3-0,2=0,1\left(mol\right)\)
⇒ mX = mCu + mCuO dư = 0,2.64 + 0,1.80 = 20,8 (g)
a) \(n_{Al}=\dfrac{10,8}{27}=0,4\left(mol\right)\)
PTHH: 2Al + 6HCl --> 2AlCl3 + 3H2
0,4-------------->0,4-->0,6
=> \(m_{AlCl_3}=0,4.133,5=53,4\left(g\right)\)
b) \(V_{H_2}=0,6.22,4=13,44\left(l\right)\)
c) \(n_{CuO}=\dfrac{8}{80}=0,1\left(mol\right)\)
PTHH: CuO + H2 --to--> Cu + H2O
Xét tỉ lệ: \(\dfrac{0,1}{1}< \dfrac{0,6}{1}\) => CuO hết, H2 dư
PTHH: CuO + H2 --to--> Cu + H2O
0,1------------>0,1
=> mchất rắn = 0,1.64 = 6,4 (g)
\(n_{Al}=\dfrac{10,8}{27}=0,4\left(mol\right)\\ pthh:2Al+6HCl\rightarrow2AlCl_3+3H_2\)
Mol : 0,4 0,4 0,6
\(m_{AlCl_3}=133,5.0,4=53,4\left(g\right)\\ V_{H_2}=0,6.22,4=13,44\left(l\right)\)
\(n_{CuO}=\dfrac{8}{80}=0,8\left(mol\right)\\
pthh:CuO+H_2\underrightarrow{t^o}Cu+H_2O\\
LTL:\dfrac{0,8}{1}>\dfrac{0,6}{1}\)
=> CuO dư
\(n_{CuO\left(p\text{ư}\right)}=n_{Cu}=n_{H_2}=0,6\left(mol\right)\\
m_{CuO\left(d\right)}=\left(0,8-0,6\right).80=16\left(g\right)\\
m_{Cu}=0,6.64=38,4\left(g\right)\\
m_{cr}=16+38,4=54,4\left(g\right)\)
\(n_{Mg}=\dfrac{7,2}{24}=0,3\left(mol\right)\\
pthh:Mg+2HCl\rightarrow MgCl_2+H_2\)
0,3 0,3 0,3
\(m_{MgCl_2}=0,3.95=28,5g\\
V_{H_2}=0,3.22,4=6,72l\\
n_{CuO}=\dfrac{3}{80}=0,0375\left(mol\right)\\
pthh:CuO+H_2\underrightarrow{t^o}Cu+H_2O\\
LTL:\dfrac{0,0375}{1}>\dfrac{0,3}{1}\)
=>Hidro dư
\(n_{Cu}=n_{CuO}=0,0375\left(mol\right)\\
m_{Cu}=0,0375.64=2,4\left(g\right)\)
a) 2Al+6HCl→2AlCl3+3H22Al+6HCl→2AlCl3+3H2
b) nAl=5,427=0,2(mol)nAl=5,427=0,2(mol)
Theo phương trình : nH2=32nAl=0,3(mol)nH2=32nAl=0,3(mol)
→VH2(đktc)=0,3.22,4=6,72(l)→VH2(đktc)=0,3.22,4=6,72(l)
c) Chất rắn : 0,2(mol)0,2(mol)
CuO dư : 0,2(mol)Cu0,2(mol)Cu
%CuO=0,2.80(0,2.80+0,2.64).100=55,56%%CuO=0,2.80(0,2.80+0,2.64).100=55,56%
%Cu=44,44%%Cu=44,44%
a)\(n_{Al}=\dfrac{5,4}{27}=0,2mol\)
\(2Al+3H_2SO_4\rightarrow Al_2\left(SO_4\right)_3+3H_2\)
0,2 0,3 0,1 0,3
b)\(V_{H_2}=0,3\cdot22,4=6,72l\)
c)\(n_{CuO}=\dfrac{32}{80}=0,4mol\)
\(CuO+H_2\rightarrow Cu+H_2O\)
0,4 0,3 0,3
\(m_{Cu}=0,3\cdot64=19,2g\)
\(n_{HCl}=0,4.1=0,4\left(mol\right)\\ pthh:Zn+2HCl\rightarrow ZnCl_2+H_2\uparrow\)
0,2 0,4 0,2
\(m_{Zn}=0,2.65=13\left(g\right)\\ V_{H_2}=0,2.22,4=4,48\left(l\right)\\ n_{CuO}=\dfrac{20}{80}=0,25\left(mol\right)\\ pthh:CuO+H_2\underrightarrow{t^o}Cu+H_2O\\ LTL:\dfrac{0,25}{1}>\dfrac{0,2}{1}\)
=> CuO dư
\(n_{CuO\left(p\text{ư}\right)}=n_{Cu}=n_{H_2}=0,2\left(mol\right)\\ X=\left\{{}\begin{matrix}m_{CuO\left(d\right)}=\left(0,25-0,2\right).80=4\left(g\right)\\m_{Cu}=0,2.64=12,8\left(g\right)\end{matrix}\right.=4+12,8=16,8\left(g\right)\)