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\(2Al+3H_2SO_4\rightarrow Al_2\left(SO_4\right)_3+3H_2\)
a) \(n_{Al}=\dfrac{10,8}{27}=0,4\left(mol\right)\)
Theo pt: \(n_{H_2}=\dfrac{3}{2}n_{Al}=\dfrac{3}{2}.0,4=0,6\left(mol\right)\)
\(V_{H_2}=0,6.22,4=13,44\left(l\right)\)
b) Theo pt: \(n_{H_2SO_4}=\dfrac{3}{2}n_{Al}=\dfrac{3}{2}.0,4=0,6\left(mol\right)\)
\(m_{H_2SO_4}=0,6.98=58,8g\)
\(C_{\%}dd_{H_2SO_4}=\dfrac{m_{ct}}{m_{dd}}.100\%=\dfrac{58,8}{200}.100\%=29,4\%\)
c) Theo pt: \(n_{Al_2\left(SO_4\right)_3}=\dfrac{n_{Al}}{2}=0,2\left(mol\right)\)
\(m_{Al_2\left(SO_4\right)_3}=0,2.342=68,4\left(g\right)\)
Áp dụng định luật bảo toàn khối lượng
\(m_{dd_{Al_2\left(SO_4\right) _3}}=m_{Al}+m_{dd_{H_2SO_4}}-m_{H_2}\)
\(=10,8+200-0,6.2=209,6g\)
\(C_{\%_{Al_2\left(SO_4\right)_3}}=\dfrac{68,4}{209,6}.100\%\approx32,6\%\)
\(a) n_{Zn} = \dfrac{19,5}{65} = 0,3(mol) ; n_{HCl} = 0,35.2 = 0,7(mol)\\ Zn + 2HCl \to ZnCl_2 + H_2\\ n_{HCl} = 0,7 > 2n_{Zn} = 0,6 \to HCl\ dư\\ n_{H_2} = n_{Zn} = 0,3(mol) \Rightarrow V_{H_2} = 0,3.22,4 = 6,72(lít)\\ b) m_{dd\ HCl} = 350.1,05 = 367,5(gam)\\ m_{dd\ sau\ pư} = 19,5 + 367,5 - 0,3.2 = 386,4(gam)\\ \Rightarrow C\%_{ZnCl_2} = \dfrac{0,3.136}{386,4}.100\% = 10,56\%\\ c) C\%_{HCl} = \dfrac{0,7.36,5}{367,5}.100\% = 6,95\%\)
a, \(Zn+2HCl\rightarrow ZnCl_2+H_2\)
Ta có: \(n_{Zn}=\dfrac{13}{65}=0,2\left(mol\right)\)
Theo PT: \(n_{ZnCl_2}=n_{H_2}=n_{Zn}=0,2\left(mol\right)\)
\(\Rightarrow V_{H_2}=0,2.24,79=4,958\left(l\right)\)
b, \(n_{HCl}=2n_{Zn}=0,4\left(mol\right)\Rightarrow V_{HCl}=\dfrac{0,4}{2}=0,2\left(l\right)\)
c, \(C_{M_{ZnCl_2}}=\dfrac{0,2}{0,2}=1\left(M\right)\)
a) \(n_{Na_2CO_3}=\dfrac{21,2}{106}=0,2\left(mol\right)\)
PTHH: Na2CO3 + 2HCl --> 2NaCl + CO2 + H2O
0,2------------------>0,4---->0,2
mdd sau pư = 200 + 21,2 - 0,2.44 = 212,4(g)
=> \(C\%\left(NaCl\right)=\dfrac{0,4.58,5}{212,4}.100\%=11,017\%\)
$PTHH:Na_2CO_3+2HCl\to 2NaCl+H_2O+CO_2\uparrow$
$n_{Na_2CO_3}=\dfrac{21,2}{106}=0,2(mol)$
Theo PT: $n_{NaCl}=n_{CO_2}=0,2(mol)$
$\Rightarrow m_{NaCl}=0,4.58,5=23,4(g);m_{CO_2}=0,2.44=8,8(g)$
$\Rightarrow C\%_{NaCl}=\dfrac{23,4}{21,2+200-8,8}.100\%\approx 11,01\%$
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) PTHH: \(Zn+2HCl\rightarrow ZnCl_2+H_2\uparrow\)
b) Ta có: \(n_{H_2}=\dfrac{1,12}{22,4}=0,05\left(mol\right)\)
\(\Rightarrow n_{HCl}=0,1mol\) \(\Rightarrow C_{M_{HCl}}=\dfrac{0,1}{0,2}=0,5\left(M\right)\)
c) Theo PTHH: \(n_{Zn}=n_{H_2}=0,05mol\)
\(\Rightarrow m_{Zn}=0,05\cdot65=3,25\left(g\right)\)
\(\Rightarrow\%m_{Zn}=\dfrac{3,25}{8,37}\cdot100\%\approx38,83\%\) \(\Rightarrow\%m_{Cu}=61,17\%\)
