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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)
\(n_{Al}=\dfrac{5,4}{27}=0,2mol\)
a)\(2Al+6HCl\rightarrow2AlCl_3+3H_2\)
0,2 0,6 0,3
\(C_M=\dfrac{0,6}{0,4}=1,5M\)
b)\(n_{CuO}=\dfrac{32}{80}=0,4mol\)
\(CuO+H_2\rightarrow Cu+H_2O\)
0,4 0,3 0,3
Sau phản ứng CuO dư và dư \(\left(0,4-0,3\right)\cdot80=8g\)
\(m_{rắn}=m_{Cu}=0,3\cdot64=19,2g\)
a) nKMnO4=0,01(mol)
PTHH: 2 KMnO4 -to-> K2MnO4 + MnO2 + O2
0,01______________0,005_____0,005___0,005(mol)
V(O2,đktc)=0,005.22,4=0,112(l)
b) PTHH: 2 Cu + O2 -to-> 2 CuO
nCu=0,1(mol); nO2=0,005(mol)
Ta có: 0,1/2 > 0,005/1
=> Cu dư, O2 hết, tính theo nO2.
nCu(p.ứ)=2.0,005=0,01(mol)
=> nCu(dư)=0,1-0,01=0,09(mol)
=>mCu(dư)=0,09.64=5,76(g)
a. \(n_P=\frac{6,2}{31}=0,2mol\)
\(V_{O_2}=V_{kk}.\frac{1}{5}=\frac{18,48}{5}=3,696l\)
\(n_{O_2}=\frac{3,696}{22,4}=0,165mol\)
PTHH: \(4P+5O_2\xrightarrow{t^o}2P_2O_5\)
Tỷ lệ \(\frac{0,2}{4}>\frac{0,165}{5}\)
Vậy P dư
\(n_{P\left(\text{phản ứng }\right)}=\frac{4}{5}n_{O_2}=0,132mol\)
\(n_{P\left(dư\right)}=0,2-0,132=0,068mol\)
\(\rightarrow m_{P\left(dư\right)}=0,068.31=2,108g\)
b. \(n_{P_2O_5}=\frac{2}{5}n_{O_2}=0,066mol\)
\(\rightarrow m_{P_2O_5}=0,066.142=9,372g\)
c. PTHH: \(2KClO_3\xrightarrow{t^o}2KCl+3O_2\)
\(n_{KClO_3}=\frac{2}{3}n_{O_2}=0,11mol\)
\(\rightarrow m_{KClO_3}=0,11.122,5=13,475g\)
PTHH: \(2Al+3H_2SO_4\rightarrow Al_2\left(SO_4\right)_3+3H_2\uparrow\)
Làm gộp các phần còn lại
Ta có: \(n_{Al}=\dfrac{5,4}{27}=0,2\left(mol\right)\) \(\Rightarrow\left\{{}\begin{matrix}n_{Al_2\left(SO_4\right)_3}=0,1mol\\n_{H_2SO_4}=n_{H_2}=0,3mol\end{matrix}\right.\)
\(\Rightarrow\left\{{}\begin{matrix}V_{H_2}=0,3\cdot22,4=6,72\left(l\right)\\m_{Al_2\left(SO_4\right)_3}=0,1\cdot342=34,2\left(g\right)\\m_{H_2SO_4}=0,3\cdot98=29,4\left(g\right)\end{matrix}\right.\)
\(n_{Zn}=\dfrac{m}{M}=\dfrac{3,25}{65}=0,05\left(mol\right)\)
\(PTHH:Zn+2HCl\rightarrow ZnCl_2+H_2\)
1 2 1 1
0,05 0,1 0,05 0,05
a) \(V_{H_2}=n.24,79=0,05.24,79=1,2395\left(l\right)\)
\(m_{ZnCl_2}=n.M=0,05.\left(65+35,5.2\right)=6,8\left(g\right)\)
b) \(PTHH:CuO+H_2\underrightarrow{t^o}Cu+H_2O\)
Ta cos tỉ lệ: \(\dfrac{0,1}{1}>\dfrac{0,05}{1}\Rightarrow\) CuO dư.
Theo ptr, ta có: \(n_{Cu}=n_{H_2}=0,05mol\\ \Rightarrow m_{Cu}=n.M=0,05.64=3,2\left(g\right).\)
a, PT: \(Zn+2HCl\rightarrow ZnCl_2+H_2\)
Ta có: \(n_{Zn}=\dfrac{3,25}{65}=0,05\left(mol\right)\)
THeo PT: \(n_{ZnCl_2}=n_{H_2}=n_{Zn}=0,05\left(mol\right)\)
\(\Rightarrow V_{H_2}=0,05.24,79=1,2395\left(l\right)\)
\(m_{ZnCl_2}=0,05.136=6,8\left(g\right)\)
b, Ta có: \(n_{CuO}=\dfrac{8}{80}=0,1\left(mol\right)\)
PT: \(CuO+H_2\underrightarrow{t^o}Cu+H_2O\)
Xét tỉ lệ: \(\dfrac{0,1}{1}>\dfrac{0,05}{1}\), ta được CuO dư.
Theo PT: \(n_{Cu}=n_{H_2}=0,05\left(mol\right)\Rightarrow m_{Cu}=0,05.64=3,2\left(g\right)\)
