PTHH: \(Fe_2O_3+6HCl\rightarrow2FeCl_3+3H_2O\)

a+b) Ta có: \(n_{Fe_2O_3}=\dfrac{16}{160}=0,1\left(mol\right)\)

\(\Rightarrow\left\{{}\begin{matrix}n_{HCl}=0,6\left(mol\right)\\n_{FeCl_3}=0,2\left(mol\right)\end{matrix}\right.\) \(\Rightarrow\left\{{}\begin{matrix}m_{ddHCl}=\dfrac{0,6\cdot36,5}{14,6\%}=150\left(g\right)\\m_{FeCl_3}=0,2\cdot162,5=32,5\left(g\right)\end{matrix}\right.\)

\(\Rightarrow C\%_{FeCl_3}=\dfrac{32,5}{150+16}\cdot100\%\approx19,58\%\)

b) PTHH: \(HCl+KOH\rightarrow KCl+H_2O\)

Theo PTHH: \(n_{KOH}=n_{HCl}=0,6\left(mol\right)\) \(\Rightarrow V_{KOH}=\dfrac{0,6}{0,5}=1,2\left(l\right)\)

\(n_{Na_2O}=\dfrac{12,4}{62}=0,2mol\\ a,Na_2O+H_2O\rightarrow2NaOH\\ NaOH+HCl\rightarrow NaCl+H_2O\\ n_{HCl}=n_{NaCl}=0,2.2=0,4mol\\ b,C_{M_{HCl}}=\dfrac{0,4}{0,2}=2M\\ c,m_{ddNaCl}=200.1,05+12,4=222,4g\\ C_{\%NaCl}=\dfrac{0,4.58,5}{222,4}\cdot100=10,52\%\)

a) $n_{Al} = \dfrac{5,4}{27} = 0,2(mol)$
$2Al + 6HCl \to 2AlCl_3 + 3H_2$
$n_{HCl} = 3n_{Al} = 0,6(mol)$
$m_{HCl} = 0,6.36,5 = 21,9(gam)$

b) $n_{H_2} = \dfrac{3}{2}n_{Al} = 0,3(mol)$
$V_{H_2} = 0,3.22,4 = 6,72(lít)$

c)

$m_{dd\ sau\ pư} = 5,4 + 200 - 0,3.2 = 204,8(gam)$

$m_{HCl\ dư} = 200.20\% - 21,9 = 18,1(gam)$

$C\%_{HCl} = \dfrac{18,1}{204,8}.100\% = 8,84\%$
$C\%_{AlCl_3} = \dfrac{0,2.133,5}{204,8}.100\% = 13,04\%$

a) nH2=32nAl=0,3(mol)nH2=32nAl=0,3(mol)
VH2=0,3.22,4=6,72(lít)VH2=0,3.22,4=6,72(lít)

c)

mdd sau pư=5,4+200−0,3.2=204,8(gam)

mHCl dư=200.20%−21,9=18,1(gam)mHCl dư=200.20%−21,9=18,1(gam)

C%AlCl3=0,2.133,5204,8.100%=13,04%

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)

=> \(C\%ddAlCl3=\dfrac{66,75}{312}.100\approx21,394\%\)

\(a) n_{Zn} = \dfrac{19,5}{65} = 0,3(mol) ; n_{HCl} = 0,35.2 = 0,7(mol)\\ Zn + 2HCl \to ZnCl_2 + H_2\\ n_{HCl} = 0,7 > 2n_{Zn} = 0,6 \to HCl\ dư\\ n_{H_2} = n_{Zn} = 0,3(mol) \Rightarrow V_{H_2} = 0,3.22,4 = 6,72(lít)\\ b) m_{dd\ HCl} = 350.1,05 = 367,5(gam)\\ m_{dd\ sau\ pư} = 19,5 + 367,5 - 0,3.2 = 386,4(gam)\\ \Rightarrow C\%_{ZnCl_2} = \dfrac{0,3.136}{386,4}.100\% = 10,56\%\\ c) C\%_{HCl} = \dfrac{0,7.36,5}{367,5}.100\% = 6,95\%\)

\(n_{HCl}=\dfrac{100.7,3\%}{36,5}=0,2\left(mol\right)\\ a,PTHH:Zn+2HCl\rightarrow ZnCl_2+H_2\\ b,n_{Zn}=n_{H_2}=n_{ZnCl_2}=\dfrac{0,2}{2}=0,1\left(mol\right)\\ m=m_{Zn}=0,1.65=6,5\left(g\right)\\ c,V_{H_2\left(đktc\right)}=0,1.22,4=2,24\left(l\right)\\ d,m_{ddZnCl_2}=6,5+100-0,1.2=106,3\left(g\right)\\ C\%_{ddZnCl_2}=\dfrac{0,1.136}{106,3}.100\approx12,794\%\)

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

\(n_{Zn}=\dfrac{26}{65}=0,4\left(mol\right)\)

Pt : \(Zn+2HCl\rightarrow ZnCl_2+H_2|\)

        1          2             1          1

       0,4       0,8          0,4       0,4

a) \(n_{H2}=\dfrac{0,4.1}{1}=0,4\left(mol\right)\)

\(V_{H2\left(dktc\right)}=0,4.22,4=8,96\left(l\right)\)

b) \(n_{HCl}=\dfrac{0,4.2}{1}=0,8\left(mol\right)\)

⇒ \(m_{HCl}=0,8.36,5=29,2\left(g\right)\)

\(m_{ddHCl}=\dfrac{29,2.100}{14,6}=200\left(g\right)\)

c) \(n_{ZnCl2}=\dfrac{0,4.1}{1}=0,4\left(mol\right)\)

⇒ \(m_{ZnCl2}=0,4.136=54,4\left(g\right)\)

\(m_{ddspu}=26+200-\left(0,4.2\right)=225,2\left(g\right)\)

\(C_{ZnCl2}=\dfrac{54,4.100}{225,2}=24,16\)⁰/₀

 Chúc bạn học tốt

\(n_{CuSO4}=\dfrac{16\%.50}{100\%.160}=0,05\left(mol\right)\)

Pt : \(CuSO_4+2NaOH\rightarrow Cu\left(OH\right)_2+Na_2SO_4\)

       0,05-------->0,1---------->0,05--------->0,05

a) \(C\%_{ddNaOH}=\dfrac{0,1.40}{250}.100\%=1,6\%\)

b) \(m_{ddspu}=50+250-0,05.98=295,1\left(g\right)\)

\(C\%_{Na2SO4}=\dfrac{0,05.142}{295,1}.100\%=2,41\%\)

\(n_{CuSO_4}=\dfrac{m_{dd}\cdot C\%}{100\cdot M}=\dfrac{50\cdot16\%}{100\cdot\left(64+32+16\cdot4\right)}=0,05\left(mol\right)\)

\(PTHH:CuSO_4+2NaOH\rightarrow Cu\left(OH\right)_2+Na_2SO_4\)

                     1                  2                    1                    1

                   0,05              0,1                 0,05               0,05 (mol)

\(a)C\%_{NaOH}=\dfrac{n\cdot100\cdot M}{m_{dd}}=\dfrac{01\cdot100\cdot\left(23+16+1\right)}{250}=1,6\%\)

\(b)m_{dd-sau-pư}=m_{dd_đ}+m_{ct_đ}-m\downarrow-m\uparrow\)

\(=50+250-\left(0,05\cdot23+32+16\cdot4\right)=294,05\left(g\right)\)

\(C\%_{Na_2SO_4}=\dfrac{n\cdot100\cdot M}{m_{dd}}=\dfrac{0,05\cdot100\cdot\left(23\cdot2+32+16\cdot4\right)}{294,05}\approx2,41\%.\)

nCuSO4​=100%.16016%.50​=0,05(mol)

Pt : ����4+2����→��(��)2+��2��4CuSO4​+2NaOH→Cu(OH)2​+Na2​SO4​

       0,05-------->0,1---------->0,05--------->0,05

a) �%������=0,1.40250.100%=1,6%C%ddNaOH​=2500,1.40​.100%=1,6%

b) ������=50+250−0,05.98=295,1(�)mddspu​=50+250−0,05.98=295,1(g)

�%��2��4=0,05.142295,1.100%=2,41%C%Na2SO4​=295,10,05.142​.100%=2,41%