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\(PTHH:4Al+6HCl\rightarrow2Al_2Cl_3+3H_2\uparrow\)
\(n_{Al}=\frac{3,78}{27}=0,14\left(mol\right)\)
\(\Rightarrow n_{H_2}=\frac{3}{4}n_{Al}=0,105\left(mol\right)\)
\(V_{H_2}=0,105.22,4=2,352\left(l\right)\)
\(n_{HCl}=\frac{3}{2}n_{Al}=\frac{3}{2}.0,14=0,21\left(mol\right)\)
\(C_{M_{ddHCl}}=\frac{0,21}{0,2}=1,05\left(M\right)\)
\(n_{Al_2Cl_3}=\frac{1}{2}n_{Al}=\frac{1}{2}.0,14=0,07\left(mol\right)\)
\(m_{Al_2Cl_3}=0,07.160,5=11,235\left(g\right)\)
\(n_{K2O}=\dfrac{9,4}{94}=0,1\left(mol\right)\)
Pt : \(K_2O+H_2O\rightarrow2KOH|\)
1 1 2
0,1 0,2
a) \(n_{KOH}=\dfrac{0,1.2}{1}=0,2\left(mol\right)\)
200ml = 0,2l
\(C_{M_{ddKOH}}=\dfrac{0,2}{0,2}=1\left(M\right)\)
b) Pt : \(KOH+HCl\rightarrow KCl+H_2O|\)
1 1 1 1
0,1 0,1
\(n_{HCl}=\dfrac{0,1.1}{1}=0,1\left(mol\right)\)
\(m_{HCl}=0,1.36,5=3,65\left(g\right)\)
\(m_{ddHCl}=\dfrac{3,65.100}{10}=36,5\left(g\right)\)
c) \(CO_2+2KOH\rightarrow K_2CO_3+H_2O|\)
1 2 1 1
0,05 0,1
\(n_{CO2}=\dfrac{0,1.1}{2}=0,05\left(mol\right)\)
\(V_{CO2\left(dktc\right)}=0,05.22,4=1,12\left(l\right)\)
Chúc bạn học tốt
Ta có: \(n_{Mg}=\dfrac{4,8}{24}=0,2\left(mol\right)\)
PT: \(Mg+2HCl\rightarrow MgCl_2+H_2\)
a, \(n_{H_2}=n_{Mg}=0,2\left(mol\right)\)
\(\Rightarrow V_{H_2}=0,2.22,4=4,48\left(l\right)\)
b, \(n_{HCl}=2n_{H_2}=0,4\left(mol\right)\)
\(\Rightarrow C_{M_{HCl}}=\dfrac{0,4}{0,2}=2\left(M\right)\)
c, PT: \(HCl+KOH\rightarrow KCl+H_2O\)
Theo PT: \(n_{KOH}=n_{HCl}=0,4\left(mol\right)\)
\(\Rightarrow m_{ddKOH}=\dfrac{0,4.56}{5,6\%}=400\left(g\right)\)
\(\Rightarrow V_{ddKOH}=\dfrac{400}{1,045}\approx382,78\left(ml\right)\)
nAl= 0,5(mol)
a) PTHH: 2 Al + 6 HCl -> 2 AlCl3 + 3 H2
nHCl= 6/2 . 0,5= 1,5(mol)
=>mHCl= 1,5.36,5=54,75(mol)
=> mddHCl= (54,75.100)/18,25=300(g)
b) nH2= 3/2. 0,5=0,75(mol)
=>V(H2,đktc)=0,75.22,4=16,8(l)
c) nAlCl3= nAl= 0,5(mol) -> mAlCl3=0,5. 133,5=66,75(g)
mddAlCl3=mAl+ mddHCl - mH2= 13,5 + 300-0,75.2=312(g)
=> \(C\%ddAlCl3=\dfrac{66,75}{312}.100\approx21,394\%\)
a, \(n_{H_2}=\dfrac{3,36}{22,4}=0,15\left(mol\right)\)
\(Fe+2HCl\rightarrow FeCl_2+H_2\)
\(n_{Fe}=n_{H_2}=0,15\left(mol\right)\Rightarrow m_{Fe}=0,15.56=8,4\left(g\right)\)
b, \(n_{Na_2O}=\dfrac{6,2}{62}=0,1\left(mol\right)\)
\(Na_2O+H_2O\rightarrow2NaOH\)
\(n_{NaOH}=2n_{Na_2O}=0,2\left(mol\right)\Rightarrow C_{M_{NaOH}}=\dfrac{0,2}{0,5}=0,4\left(M\right)\)
\(a,n_{H_2}=\dfrac{3,36}{22,4}=0,15\left(mol\right)\)
PTHH:
\(Fe+2HCl\rightarrow FeCl_2+H_2\uparrow\)
0,15 0,3 0,15 0,15
\(m_{Fe}=0,15.56=8,4\left(g\right)\)
\(a,n_{Na_2O}=\dfrac{6,2}{62}=0,1\left(mol\right)\)
PTHH :
\(Na_2O+H_2O\rightarrow2NaOH\)
0,1 0,1 0,2
\(C_{M\left(A\right)}=\dfrac{0,2}{0,5}=0,4\left(M\right)\)
\(a,n_{Mg}=\dfrac{4,8}{24}=0,2\left(mol\right)\\ Mg+2HCl\rightarrow MgCl_2+H_2\\ b,n_{MgCl_2}=n_{H_2}=n_{Mg}=0,2\left(mol\right)\\ b,V_{H_2\left(25\text{đ}\text{ộ}C,1bar\right)}=0,2.24,79=4,958\left(l\right)\\ c,n_{HCl}=2.0,2=0,4\left(mol\right)\\ m_{\text{dd}HCl}=\dfrac{0,4.36,5.100}{10}=146\left(g\right)\\ m_{\text{dd}A}=m_{Mg}+m_{\text{dd}HCl}-m_{H_2}=4,8+146-0,2.2=150,4\left(g\right)\\ d,C\%_{\text{dd}MgCl_2}=\dfrac{95.0,2}{150,4}.100\approx12,633\%\)