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nAl = 5.4 / 27 = 0.2 (mol)
2Al + 6HCl => 2AlCl3 + 3H2
0.2......0.6............0.2.......0.3
a) VH2 = 0.3 * 22.4 = 6.72 (l)
b) mAlCl3 = 0.2 * 133.5 = 26.7 (g)
c) VddHCl = 0.6 / 1.5 = 0.4 (l)
d) CMAlCl3 = 0.2 / 0.4 = 0.5 (M)
PTHH: \(2Al+6HCl\rightarrow2AlCl_3+3H_2\uparrow\)
Ta có: \(n_{Al}=\dfrac{5,4}{27}=0,2\left(mol\right)\)
\(\Rightarrow\left\{{}\begin{matrix}n_{HCl}=0,6\left(mol\right)\\n_{AlCl_3}=0,2\left(mol\right)\\n_{H_2}=0,3\left(mol\right)\end{matrix}\right.\) \(\Rightarrow\left\{{}\begin{matrix}V_{H_2}=0,3\cdot22,4=6,72\left(l\right)\\m_{AlCl_3}=0,2\cdot133,5=26,7\left(g\right)\\V_{HCl}=\dfrac{0,6}{1,5}=0,4\left(l\right)=400\left(ml\right)\\C_{M_{AlCl_3}}=\dfrac{0,2}{0,4}=0,5\left(M\right)\end{matrix}\right.\)
a) K2CO3 + 2HCl --> 2KCl + CO2 + H2O
b) \(n_{K_2CO_3}=\dfrac{13,8}{138}=0,1\left(mol\right)\)
PTHH: K2CO3 + 2HCl --> 2KCl + CO2 + H2O
______0,1----->0,2------>0,2--->0,1
=> VCO2 = 0,1.22,4 = 2,24 (l)
c) mHCl = 0,2.36,5 = 7,3 (g)
\(m_{ddHCl}=\dfrac{7,3.100}{7,3}=100\left(g\right)\)
d) mKCl = 0,2.74,5 = 14,9 (g)
mdd sau pư = 13,8 + 100 - 0,1.44 = 109,4 (g)
=> \(C\%\left(KCl\right)=\dfrac{14,9}{109,4}.100\%=13,62\%\)
đổi 200 ml = 0,02 l
a) PTHH : Fe + HCl -> FeCl2 + H2
b) \(n_{Fe}=\dfrac{5,6}{56}=0,1\left(mol\right)\)
\(=>V_{H_2}=0,1.22,4=2,24\left(l\right)\)
\(C_{HCl}=\dfrac{n}{V}=\dfrac{0,1}{0,02}=5\left(M\right)\)
Đông Hải làm câu 1 rồi thì tui làm phần còn lại
Câu 2:
\(a,PTHH:CuO+H_2\underrightarrow{t^o}Cu+H_2O\\ b,n_{CuO}=\dfrac{m}{M}=\dfrac{8}{80}=0,1\left(mol\right)\\ Theo.PTHH:n_{Cu}=n_{H_2}=n_{CuO}=0,1\left(mol\right)\\ V_{H_2\left(đktc\right)}=n.22,4=0,1.22,4=2,24\left(l\right)\\ c,m_{Cu}=n.M=0,1.64=6,4\left(g\right)\\ d,PTHH:2H_2+O_2\underrightarrow{t^o}2H_2O\left(2\right)\\ Theo.PTHH\left(2\right):n_{H_2O}=n_{H_2}=0,1\left(mol\right)\)
\(m_{H_2O}=n.M=0,1.18=1,8\left(g\right)\)
nZn= 13/65=0,2(mol)
a) PTHH: Zn + 2 HCl -> ZnCl2 + H2
b) nH2=nZnCl2=nZn=0,2(mol)
=>V(H2,đktc)=0,2 x 22,4= 4,48(l)
c) khối lượng muối sau phản ứng chứ nhỉ?
mZnCl2=136.0,2=27,2(g)
Bài 1:
A . PTHH : 2AI + 3H2SO4 => AI2(SO4)3 + 3H2
B . nAI = 7.1 : 27 = 0.2 (mol)
PT :2AI + 3H2SO4 => AI2(SO4)3 + 3H2
2 mol 3 mol 3 mol
0.2 mol 0.3 mol 0.3 mol
=> VH2 = 22.4 X 0.3 = 6.72 (lít)
3. 500ml = 0.5 lít
Nồng độ mol/l của dd H2SO4 là:
CM H2SO4 = mol/lít =0.3/0.5 = 0.6 MOL/ LÍT
Bài 8:
\(n_{H_2}=\dfrac{7,437}{24,79}=0,3\left(mol\right)\)
PT: \(2Al+6HCl\rightarrow2AlCl_3+3H_2\)
_____0,2______0,6_____0,2____0,3 (mol)
a, \(m_{Al}=0,2.27=5,4\left(g\right)\)
b, \(C_{M_{HCl}}=\dfrac{0,6}{0,3}=2\left(M\right)\)
c, \(C_{M_{AlCl_3}}=\dfrac{0,2}{0,2}=1\left(M\right)\)
Bài 9:
Ta có: \(n_{H_2}=\dfrac{2,479}{24,79}=0,1\left(mol\right)\)
PT: \(Mg+2HCl\rightarrow MgCl_2+H_2\)
\(MgO+2HCl\rightarrow MgCl_2+H_2O\)
a, \(n_{Mg}=n_{H_2}=0,1\left(mol\right)\Rightarrow m_{Mg}=0,1.24=2,4\left(g\right)\)
\(\Rightarrow m_{MgO}=8,4-2,4=6\left(g\right)\)
b, \(n_{MgO}=\dfrac{6}{40}=0,15\left(mol\right)\)
Theo PT: \(n_{HCl}=2n_{Mg}+2n_{MgO}=0,5\left(mol\right)\)
\(\Rightarrow m_{ddHCl}=\dfrac{0,5.36,5}{3,65\%}==500\left(g\right)\)
\(n_{Zn}=\dfrac{13}{65}=0,2\left(mol\right)\\
pthh:Zn+2HCl\rightarrow ZnCl_2+H_2\)
0,2 0,4 0,2
\(V_{H_2}=0,2.22,4=4,48\left(l\right)\\
C\%_{HCl}=\dfrac{0,4.36,5}{200}.100\%=7,3\%\)
\(a.BTNT\left(H\right):n_{HCl}=2n_{H_2}=0,65\left(mol\right)\\ \Rightarrow CM_{HCl}=\dfrac{0,65}{0,5}=1,3M\\ b.2Al+6HCl\rightarrow2AlCl_3+3H_2\\ Fe+2HCl\rightarrow FeCl_2+H_2\\ Đặt:\left\{{}\begin{matrix}n_{Al}=x\left(mol\right)\\n_{Fe}=y\left(mol\right)\end{matrix}\right.\\ \Rightarrow\left\{{}\begin{matrix}\dfrac{3}{2}x+y=0,325\\27x+56y=9,65\end{matrix}\right.\\ \Rightarrow\left\{{}\begin{matrix}x=0,15\\y=0,1\end{matrix}\right.\\ \Rightarrow\left\{{}\begin{matrix}m_{Al}=4,05\left(g\right)\\m_{Fe}=5,6\left(g\right)\end{matrix}\right.\)
a) Zn + 2HCl --> ZnCl2 + H2
b) \(n_{Zn}=\dfrac{13}{65}=0,2\left(mol\right)\)
PTHH: Zn + 2HCl --> ZnCl2 + H2
0,2-------------->0,2--->0,2
=> \(m_{ZnCl_2}=0,2.136=27,2\left(g\right)\)
c)
Lượng khí thu được bị thất thoát 20%
=> Lượng khí thu được chiếm 80% lượng khí sinh ra
=> \(V_{H_2}=0,2.22,4.80\%=3,584\left(l\right)\)
\(n_{HCl}=0,3.2=0,6\left(mol\right)\\a, 2Al+6HCl\rightarrow2AlCl_3+3H_2\\ b,n_{H_2}=\dfrac{3}{6}.0,6=0,3\left(mol\right)\\ V_{H_2\left(đktc\right)}=0,3.22,4=6,72\left(l\right)\\ c,n_{Al}=n_{AlCl_3}=\dfrac{2}{6}.0,6=0,2\left(mol\right)\\ m_{Al}=0,2.27=5,4\left(g\right)\\ d,V_{ddAlCl_3}=V_{ddHCl}=0,3\left(l\right)\\ C_{MddHCl}=\dfrac{0,2}{0,3}=\dfrac{2}{3}\left(M\right)\)