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\(a,\\ 1,\\ 2Al+6HCl\rightarrow2AlCl_3+3H_2\\ 2,\\ Zn+2HCl\rightarrow ZnCl_2+H_2\\ n_{H_2}=\dfrac{13,44}{22,4}=0,6\left(mol\right)\\ \Rightarrow n_{Al}=\dfrac{0,6.2}{3}=0,4\left(mol\right)\\ n_{Zn}=n_{H_2}=0,6\left(mol\right)\\ m_{Al}=0,4.27=10,8\left(g\right)\\ m_{Zn}=65.0,6=39\left(g\right)\\ \Rightarrow m_{Al}< m_{Zn}\\ b,Đặt:n_{Al}=n_{Zn}=1\left(mol\right)\\ \Rightarrow n_{H_2\left(1\right)}=1,5.1=1,5\left(mol\right)\\ n_{H_2\left(2\right)}=n_{Zn}=1\left(mol\right)\\ Vì:1,5>1\)
=> Cùng lấy một khối lượng kim loại Al hoặc Zn cho phản ứng thì lượng H2 sinh ra từ phản ứng có Al sẽ nhiều hơn.
a) \(2Al+6HCl\rightarrow2AlCl_3+3H_2\)
\(n_{Al}=\dfrac{m_{Al}}{M_{Al}}=\dfrac{54}{27}=2\left(mol\right)\)
Theo PTHH: \(n_{H_2}=\dfrac{2.3}{2}=3\left(mol\right)\)
\(\Rightarrow V_{H_2}=n_{H_2}.22,4=3.22,4=67,2l\)
b) \(2H_2+O_2\rightarrow2H_2O\)
\(n_{O_2}=\dfrac{m_{O_2}}{M_{O_2}}=\dfrac{30}{32}=0,94\left(mol\right)\)
Theo PTHH: \(n_{H_2O}=\dfrac{0,94.2}{1}=1,88\left(mol\right)\)
\(\Rightarrow m_{H_2O}=n_{H_2O}.M_{H_2O}=1,88.18=33,84\left(g\right)\)
a, \(Mg+2HCl\rightarrow MgCl_2+H_2\)
Ta có: \(n_{Mg}=\dfrac{7,2}{24}=0,3\left(mol\right)\)
Theo PT: \(n_{H_2}=n_{Mg}=0,3\left(mol\right)\Rightarrow V_{H_2}=0,3.22,4=6,72\left(l\right)\)
b, \(n_{Fe_2O_3}=\dfrac{48}{160}=0,3\left(mol\right)\)
PT: \(Fe_2O_3+3H_2\underrightarrow{t^o}2Fe+3H_2O\)
Xét tỉ lệ: \(\dfrac{0,3}{1}>\dfrac{0,3}{3}\), ta được Fe2O3 dư.
Theo PT: \(n_{Fe_2O_3\left(pư\right)}=\dfrac{1}{3}n_{H_2}=0,1\left(mol\right)\Rightarrow n_{Fe_2O_3\left(dư\right)}=0,3-0,1=0,2\left(mol\right)\)
\(\Rightarrow m_{Fe_2O_3}=0,2.160=32\left(g\right)\)
a) PTHH: \(Zn+2HCl\rightarrow ZnCl_2+H_2\uparrow\)
b) Ta có: \(n_{Zn}=\dfrac{97,5}{65}=1,5\left(mol\right)=n_{H_2}\)
\(\Rightarrow V_{H_2}=1,5\cdot22,4=33,6\left(l\right)\)
c) Khử 120 gam gì vậy bạn ??
a) PTHH: Zn+2HCl→ZnCl2+H2↑Zn+2HCl→ZnCl2+H2↑
b) Ta có: nZn=97,565=1,5(mol)=nH2nZn=97,565=1,5(mol)=nH2
⇒VH2=1,5⋅22,4=33,6(l)
c) ???
\(n_{Ag_2O}=\dfrac{23.2}{232}=0.1\left(mol\right)\)
\(Ag_2O+H_2\underrightarrow{^{t^0}}2Ag+H_2O\)
\(0.1......0.1.........0.2\)
\(V_{H_2}=0.1\cdot22.4=2.24\left(l\right)\)
\(m_{Ag}=0.2\cdot108=21.6\left(g\right)\)
\(4Ag+O_2\underrightarrow{^{t^0}}2Ag_2O\)
\(0.2.....0.05\)
\(V_{kk}=5V_{O_2}=5\cdot0.05\cdot22.4=5.6\left(l\right)\)
\(n_{CuO}=\dfrac{16}{80}=0.2\left(mol\right)\)
\(CuO+H_2\underrightarrow{^{t^0}}Cu+H_2O\)
\(0.2.......0.2......0.2\)
\(V_{H_2}=0.2\cdot22.4=4.48\left(l\right)\)
\(m_{Cu}=0.2\cdot64=12.8\left(g\right)\)
\(2Cu+O_2\underrightarrow{^{t^0}}2CuO\)
\(0.2......0.1\)
\(V_{kk}=5V_{O_2}=5\cdot0.1\cdot22.4=11.2\left(l\right)\)
a)2Al+6HCl ->2AlCl3+3H2
nAl=5.4/27=0.2mol
suy ra nH2=3/2*nAl=0.2 *3/2=0.3mol
suy ra VH2=0.3*22.4=6.72 l
b)C1 :nHCl =3*nAl=3*0.2=0.6 mol
suy ra mHCl=0.6*36.5=21.9 g
C2:nAlCl3=nAl=0.2 mol
suy ra mAlCl3=0.2*133.5=26.7g
Ta có :mHCl=mAlCl3-mH2-mAl=26.7+0.3*2-5.4=21.9g
PTHH: \(2Al+6HCl\rightarrow2AlCl_3+3H_2\uparrow\)
a) Ta có: \(n_{Al}=\dfrac{5,4}{27}=0,2\left(mol\right)\)
\(\Rightarrow n_{H_2}=0,3mol\) \(\Rightarrow V_{H_2}=0,3\cdot22,4=6,72\left(l\right)\)
b)
+) Cách 1:
Theo PTHH: \(n_{HCl}=3n_{Al}=0,6mol\) \(\Rightarrow m_{HCl}=0,6\cdot36,5=21,9\left(g\right)\)
+) Cách 2:
Theo PTHH: \(n_{Al}=n_{AlCl_3}=0,2mol\) \(\Rightarrow m_{AlCl_3}=0,2\cdot133,5=26,7\left(g\right)\)
Ta có: \(m_{H_2}=0,3\cdot2=0,6\left(g\right)\)
Bảo toàn khối lượng: \(m_{HCl}=m_{AlCl_3}+m_{H_2}-m_{Al}=21,9\left(g\right)\)
c) Ta có: \(n_{AlCl_3}=0,2mol\) \(\Rightarrow\left\{{}\begin{matrix}n_{Al}=0,2mol\\n_{Cl}=0,6mol\end{matrix}\right.\)
Nếu dùng a (g) Fe thì
\(Fe\left(\dfrac{a}{56}\right)+2HCl\rightarrow FeCl_2+H_2\left(\dfrac{a}{56}\right)\)
\(n_{Fe}=\dfrac{a}{56}\Rightarrow n_{H_2}=\dfrac{a}{56}\)
Nếu dùng a (g) Al thì
\(2Al\left(\dfrac{a}{27}\right)+6HCl\rightarrow2AlCl_3+3H_2\left(\dfrac{a}{18}\right)\)
\(n_{Al}=\dfrac{a}{27}\Rightarrow n_{H_2}=\dfrac{a}{27}.\dfrac{3}{2}=\dfrac{a}{18}\)
Vì \(\dfrac{a}{56}< \dfrac{a}{18}\) nên cùng 1 khối lượng thì dùng Al để điều chế H2 sẽ được nhiều hơn