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\(n_{Al}=\dfrac{10,8}{27}=0,4\left(mol\right)\\ pthh:2Al+6HCl\rightarrow2AlCl_3+3H_2\)
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
\(n_{CuO\left(p\text{ư}\right)}=n_{Cu}=n_{H_2}=0,6\left(mol\right)\\
m_{CuO\left(d\right)}=\left(0,8-0,6\right).80=16\left(g\right)\\
m_{Cu}=0,6.64=38,4\left(g\right)\\
m_{cr}=16+38,4=54,4\left(g\right)\)
a) \(n_{Mg}=\dfrac{7,2}{24}=0,3\left(mol\right)\)
PTHH: Mg + 2HCl --> MgCl2 + H2
0,3--------------->0,3--->0,3
=> \(m_{MgCl_2}=0,3.95=28,5\left(g\right)\)
b)
\(V_{H_2}=0,3.22,4=6,72\left(l\right)\)
c)
\(n_{CuO}=\dfrac{32}{80}=0,4\left(mol\right)\)
PTHH: CuO + H2 --to--> Cu + H2O
Xét tỉ lệ: \(\dfrac{0,4}{1}>\dfrac{0,3}{1}\) => CuO dư, H2 hết
PTHH: CuO + H2 --to--> Cu + H2O
0,3<--0,3------->0,3
=> mchất rắn = 32 - 0,3.80 + 0,3.64 = 27,2 (g)
\(n_{Mg}=\dfrac{7,2}{24}=0,3\left(mol\right)\\
pthh:Mg+2HCl\rightarrow MgCl_2+H_2\)
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
\(n_{Cu}=n_{CuO}=0,0375\left(mol\right)\\
m_{Cu}=0,0375.64=2,4\left(g\right)\)
\(n_{HCl}=0,4.1=0,4\left(mol\right)\\
pthh:Zn+2HCl\rightarrow ZnCl_2+H_2\uparrow\)
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
\(n_{CuO\left(p\text{ư}\right)}=n_{Cu}=n_{H_2}=0,2\left(mol\right)\\
X=\left\{{}\begin{matrix}m_{CuO\left(d\right)}=\left(0,25-0,2\right).80=4\left(g\right)\\m_{Cu}=0,2.64=12,8\left(g\right)\end{matrix}\right.=4+12,8=16,8\left(g\right)\)
\(n_{Zn}=\dfrac{6.5}{65}=0.1\left(mol\right)\)
\(Zn+2HCl\rightarrow ZnCl_2+H_2\)
\(0.1.................................0.1\)
\(Đặt:n_{CuO\left(pư\right)}=x\left(mol\right)\)
\(CuO+H_2\underrightarrow{t^0}Cu+H_2O\)
\(x............x\)
\(m_{cr}=6-80x+64x=5.2\left(g\right)\)
\(\Rightarrow x=0.05\)
\(H\%=\dfrac{0.05}{0.075}\cdot100\%=66.67\%\)
\(n_{HCl}=0,2.2=0,4\left(mol\right)\\ PTHH:Fe+2HCl\rightarrow FeCl_2+H_2\\ n_{Fe}=n_{H_2}=n_{FeCl_2}=\dfrac{0,4}{2}=0,2\left(mol\right)\\ a,m_{Fe}=0,2.56=11,2\left(g\right)\\ b,V_{H_2\left(đktc\right)}=0,2.22,4=4,48\left(l\right)\\ c,V_{ddFeCl_2}=V_{ddHCl}=0,2\left(l\right)\\ C_{MddFeCl_2}=\dfrac{0,2}{0,2}=1\left(M\right)\)
a)
n Al = 10,8/27 = 0,4(mol)
2Al + 6HCl → 2AlCl3 + 3H2
n H2 = \(\dfrac{3}{2}\)n Al = 0,6(mol)
=> V H2 = 0,6.22,4 = 13,44(lít)
b) n AlCl3 = n Al = 0,4(mol)
=> m AlCl3 = 0,4.133,5 = 53,4(gam)
c) n CuO = 16/80 = 0,2(mol)
CuO + H2 \(\xrightarrow{t^o}\) Cu + H2O
n CuO = 0,2 < n H2 = 0,6 => H2 dư
n H2 pư = n Cu = n CuO = 0,2 mol
Suy ra:
m H2 dư = (0,6 -0,2).2 = 0,8(gam)
m Cu = 0,2.64 = 12,8(gam)
a) nAl=0,4(mol)
PTHH: 2Al + 6HCl -> 2AlCl3 + 3H2
nH2= 3/2 . nAl=3/2 . 0,4=0,6(mol)
=>V(H2,đktc)=0,6 x 22,4= 13,44(l)
b) nAlCl3= nAl=0,4(mol)
=>mAlCl3=133,5 x 0,4= 53,4(g)
c) nCuO=0,2(mol)
PTHH: CuO + H2 -to-> Cu + H2O
Ta có: 0,2/1 < 0,6/1
=> H2 dư, CuO hết, tính theo nCuO
=> nH2(p.ứ)=nCu=nCuO=0,2(mol)
=>nH2(dư)=0,6 - 0,2=0,4(mol)
=> mH2(dư)=0,4. 2=0,8(g)
mCu=0,2.64=12,4(g)
cop tên ng ta nè
a) CuO+H2−to→Cu+H2OCuO+H2−to→Cu+H2O
nCuO(bđ)=1680=0,2(mol)nCuO(bđ)=1680=0,2(mol)
⇒nCuO(pứ)=0,2.80%=0,16(mol)⇒nCuO(pứ)=0,2.80%=0,16(mol)
nH2O=nCuO=0,16(mol)nH2O=nCuO=0,16(mol)
=> mH2O=0,16.18=2,88(g)mH2O=0,16.18=2,88(g)
b) nH2=0,15(mol)nH2=0,15(mol)
Lập tỉ lệ : 0,21>0,151⇒0,21>0,151⇒Sau phản ứng CuO dư
Chất rắn sau phản ứng là Cu, CuO dư
mcr=0,15.64+(0,2−0,15).80=13,6(g)mcr=0,15.64+(0,2−0,15).80=13,6(g)
c) Gọi x là số mol CuO phản ứng
mcr=(0,2−x).80+64x=13,28mcr=(0,2−x).80+64x=13,28
=> x=0,17 (mol)
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)