Bài 6:

\(n_{Fe\left(OH\right)_3}=\dfrac{21,4}{107}=0,2\left(mol\right)\)

PT: \(Fe\left(OH\right)_3+3HCl\rightarrow FeCl_3+3H_2O\)

_______0,2________0,6______0,2 (mol)

a, \(C\%_{HCl}=\dfrac{0,6.36,5}{200}.100\%=10,95\%\)

b, \(C\%_{FeCl_3}=\dfrac{0,2.162,5}{21,4+200}.100\%\approx14,68\%\)

Bài 7:

\(m_{H_2SO_4}=100.9,8\%=9,8\left(g\right)\Rightarrow n_{H_2SO_4}=\dfrac{9,8}{98}=0,1\left(mol\right)\)

PT: \(ZnO+H_2SO_4\rightarrow ZnSO_4+H_2O\)

______0,1______0,1_______0,1 (mol)

a, \(m_{ZnO}=0,1.81=8,1\left(g\right)\)

b, \(C\%_{ZnSO_4}=\dfrac{0,1.161}{8,1+100}.100\%\approx14,89\%\)

HCl k có phần trăm à

a) $n_{Fe} = \dfrac{11,2}{56} = 0,2(mol)$
$Fe + 2HCl \to FeCl_2 + H_2$
$n_{HCl} =2 n_{Fe} = 0,2.2 = 0,4(mol)$
$C\%_{HCl} = \dfrac{0,4.36,5}{200}.100\% = 7,3\%$

b) $n_{H_2} = n_{FeCl_2} = n_{Fe} = 0,2(mol)

Sau phản ứng, $m_{dd} = 11,2 + 200 - 0,2.2 = 210,8(gam)$
$C\%_{FeCl_2} = \dfrac{0,2.127}{210,8}.100\% = 12,05\%$

\(2Al+3H_2SO_4 \to Al_2(SO_4)_3+3H_2\\ n_{Al}=0,4(mol)\\ a/\\ n_{H_2}=\frac{3}{2}.0,4=0,6(mol)\\ V_{H_2}=0,6.22,4=13,44(l)\\ b/\\ n_{H_2SO_4}=n_{H_2}=0,6(mol)\\ n_{ddH_2SO_4}=\frac{0,6.98.100}{20}=294(g)\\ c/\\ n_{Al_2(SO_4)_3}=0,2(mol)\\ C\%_{Al_2(SO_4)_3}=\frac{0,2.342}{10,8+294-0,6.2}.100\%=22,52\%\)

\(a,n_{Al_2O_3}=\dfrac{20,4}{102}=0,2\left(mol\right)\\ PTHH:Al_2O_3+6HCl\rightarrow2AlCl_3+3H_2O\\ .......0,2......1,2......0,4.....0,6\left(mol\right)\\ V_{H_2O\left(đktc\right)}=0,6\cdot22,4=13,44\left(l\right)\\ b,m_{HCl}=36,5\cdot1,2=43,8\left(g\right)\\ c,m_{dd_{HCl}}=1,2\cdot36,5=43,8\left(g\right)\\ C\%_{dd_{HCl}}=\dfrac{43,8}{200}\cdot100\%=21,9\%\)

\(d,m_{AlCl_3}=0,4\cdot133,5=53,4\left(g\right)\\ m_{dd\left(sau.pứ\right)}=m_{Al_2O_3}+m_{dd_{HCl}}-m_{H_2}=20,4+200-10,8=209,6\left(g\right)\\ C\%_{dd_{AlCl_3}}=\dfrac{53,4}{209,6}\cdot100\%\approx25,48\%\)

d)\(m_{NaOH}=\dfrac{mct.C\%}{100}=\dfrac{100.20}{100}=20\left(g\right)\)=>\(n_{NaOH}=\dfrac{20}{40}=0,5\left(mol\right)\)

\(HCl+NaOH\rightarrow NaCl+H_2O\)

tl1............1.............1..............1(mol)

So sánh\(\dfrac{n_{NaOH}}{1}< \dfrac{n_{HCl}}{1}\left(\dfrac{0,5}{1}< \dfrac{1,2}{1}\right)\)

=> HCl Dư

Vậy Dung dịch sau phản ứng làm quỳ tím chuyển màu đỏ

\(n_{Al\left(OH\right)_3}=\dfrac{7,8}{78}=0,1mol\\ 2Al\left(OH\right)_3+3H_2SO_4\rightarrow Al_2\left(SO_4\right)_3+6H_2O\\ 0,1................0,15.............0,05............0,3\\ C_{\%H_2SO_4}=\dfrac{0,15.98}{300}\cdot100\%=4,9\%\\ C_{\%Al_2\left(SO_4\right)_3}=\dfrac{0,05.342}{7,8+300}\cdot100\%=5,56\%\)

\(n_{Zn}=\dfrac{26}{65}=0,4\left(mol\right)\\ pthh:Zn+2HCl\rightarrow ZnCl_2+H_2\uparrow\) 
           0,4       0,8        0,4           0,4
\(a,V_{H_2}=0,4.22,4=8,96\left(l\right)\\ b,C\%_{HCl}=\dfrac{0,8.36,5}{150}.100\%=19,5\%\\ c,m_{\text{dd}}=26+150-\left(0,4.2\right)=175,2\left(g\right)\\ C\%_{ZnCl_2}=\dfrac{0,4.136}{175,2}.100\%=31\%\)

\(n_{Al2O3}=\dfrac{10,2}{102}=0,1\left(mol\right)\)

Pt : \(Al_2O_3+6HCl\rightarrow2AlCl_3+3H_2O|\)

          1             6              2            3

         0,1         0,6            0,1 

a) \(n_{HCl}=\dfrac{0,1.6}{1}=0,6\left(mol\right)\)

\(m_{HCl}=0,6.36,5=21,9\left(g\right)\)

\(m_{ddHCl}=\dfrac{21,9.100}{10}=219\left(g\right)\)

b) \(n_{AlCl3}=\dfrac{0,6.2}{6}=0,2\left(mol\right)\)

⇒ \(m_{AlCl3}=0,2.133,5=26,7\left(g\right)\)

c) \(m_{ddspu}=10,2+219=229,2\left(g\right)\)

\(C_{AlCl3}=\dfrac{26,7.100}{229,2}=11,65\)⁰/₀ 

 Chúc bạn học tốt

nAl2O3=10.2:102=0.1(mol) 

PTHH:Al2O3+6HCl->2AlCl3+3H2O

theo pthh:nHCl:nAl2O3=6->nHCl=6*0.1=0.6(mol)

mHCl=0.6*36.5=21.9(g)

mdd HCl=21.9*100:14.6=150(g)

theo pthh:nAlCl3:nAl2O3=2->nAlCl3=0.1*2=0.2(mol)

mAlCl3=0.2*133.5=26.7(g)

mdd sau phản ứng:10.2+150=160.2

a)\(n_{Al_2O_3}=\dfrac{m}{M}=\dfrac{10,2}{102}=0,1\left(mol\right)\)

PTHH:\(Al_2O_3+6HCl\rightarrow2AlCl_3+3H_2O\)

             0,1          0,6           0,2         0,3         (mol)

\(m_{HCl}=n.M=0,6.36,5=21,9\left(g\right)\)

b)\(m_{AlCl_3}=n.M=0,2.133,5=26,7\left(g\right)\)

c)\(m_{ddHCl}=\dfrac{m_{HCl}}{C\%}.100\)=\(\dfrac{21,9}{10}.100\)=219(g)

theo định luật bảo toàn khối lượng

\(m_{Al_2O_3}+m_{ddHCl}=m_{ddAlCl_3}\)\(\Rightarrow m_{ddAlCl_3}=10,2+219\)=229,2(g)

\(C\%_{AlCl_3}=\dfrac{m_{AlCl_3}}{m_{ddAlCl_3}}.100\)=\(\dfrac{26,7}{229,2}.100\)=8,92%