Bảo toàn nguyên tố M: n_MSO4 = 0,25mol

Bảo toàn nguyên tố Cu: n_{CuSO4 dư} = 0,1 mol

=> M = 24 (Mg)

b.

a, \(MgCl_2+2NaOH\rightarrow Mg\left(OH\right)_2+2NaCl\)

\(Mg\left(OH\right)_2\underrightarrow{t^o}MgO+H_2O\)

b, \(n_{MgCl_2}=0,2.0,25=0,05\left(mol\right)\)

 Theo PT: \(n_{MgO}=n_{Mg\left(OH\right)_2}=n_{MgCl_2}=0,05\left(mol\right)\)

\(\Rightarrow m_{MgO}=0,05.40=2\left(g\right)\)

c, \(n_{NaOH}=2n_{MgCl_2}=0,1\left(mol\right)\)

\(\Rightarrow m_{ddNaOH}=\dfrac{0,1.40}{15\%}=\dfrac{80}{3}\left(g\right)\)

\(n_{Al}=\dfrac{2,7}{27}=0,1\left(mol\right)\)

PTHH: 2Al + 6HCl ---> 2AlCl₃ + 3H₂

           0,1                                    0,15

=> V_H2 = 0,15.22,4 = 3,36 (l)

\(n_{CuO}=\dfrac{32}{80}=0,4\left(mol\right)\)

PTHH: CuO + H₂ --t^o--> Cu + H₂O

LTL: \(0,4>0,15\rightarrow\) CuO dư

Theo pthh: \(n_{CuO\left(pư\right)}=n_{Cu}=n_{H_2}=0,15\left(mol\right)\)

\(\rightarrow\left\{{}\begin{matrix}\%m_{Cu}=\dfrac{0,15.64}{0,15.64+\left(0,4-0,15\right).80}=32,43\%\\\%m_{CuO}=100\%-32,43\%=67,57\%\end{matrix}\right.\)

a. \(2Al+6HCl\rightarrow2AlCl_3+3H_2\left(1\right)\)

b. \(n_{Al}=\frac{m}{M}=\frac{2,7}{27}=0,1mol\)

Theo phương trình `(1)` \(n_{H_2}=\frac{3}{2}.n_{Al}=\frac{3}{2}.0,1=0,15mol\)

\(\rightarrow V_{H_2\left(ĐKTC\right)}=n.22,4=0,15.22,4=3,36l\)

c. \(CuO+H_2\rightarrow^{t^o}Cu+H_2O\left(2\right)\)

\(n_{CuO}=\frac{m}{M}=\frac{32}{80}=0,4mol\)

Tỷ lệ \(\frac{0,4}{1}>\frac{0,15}{1}\)

`->CuO` dư

Theo phương trình `(2)` \(n_{Cu}=n_{H_2}=0,15mol\)

\(n_{CuO\left(pứ\right)}=n_{H_2}=0,15mol\)

\(\rightarrow n_{CuO\left(dư\right)}=0,4-0,15=0,25mol\)

\(m\left(g\right)\text{ chất rắn }\hept{\begin{cases}CuO_{dư}=0,25mol\\Cu=0,15mol\end{cases}}\)

\(\rightarrow m=0,15.64+0,25.80=29,6g\)

\(\%m_{CuO\left(dư\right)}=\frac{0,25.80.100}{29,6}\approx67,6\%\)

\(\%m_{Cu}=100\%-67,6\%=32,4\%\)

a. \(n_{Al}=\dfrac{2.7}{27}=0,1\left(mol\right)\)

PTHH : 2Al + 6HCl -> 2AlCl₃ + 3H₂

             0,1                                0,15

b. \(V_{H_2}=0,15.22,4=3,36\left(l\right)\)

c. \(n_{CuO}=\dfrac{32}{80}=0,4\left(mol\right)\)

PTHH : CuO + H₂ ---t^o---> Cu + H₂O

             0,15     0,15            0,15

Ta thấy : 0,4 > 0,15 =>  CuO dư , H₂ dư

\(\%CuO=\dfrac{0,15.80}{0,15.80+0,15.64}.100\%=55,55\%\)

\(\%Cu=100\%-55,55\%=44,45\%\)

Câu 3 : 

\(m_{ct}=\dfrac{10.80}{100}=8\left(g\right)\)

\(n_{NaOH}=\dfrac{8}{40}=0,2\left(mol\right)\)

a) Hiện tượng : Xuất hiện kết tủa trắng

Pt : \(2NaOH+MgSO_4\rightarrow Na_2SO_4+Mg\left(OH\right)_2|\)

           2                1                  1               1

        0,2                0,1               0,1            0,1

\(n_{Mg\left(OH\right)2}=\dfrac{0,2.1}{2}=0,1\left(mol\right)\)

⇒ \(m_{Mg\left(OH\right)2}=0,1.58=5,8\left(g\right)\)

b) \(n_{MgSO4}=\dfrac{0,2.1}{2}=0,1\left(mol\right)\)

 \(m_{MgSO4}=0,1.120=12\left(g\right)\)

\(m_{ddMgSO}=\dfrac{12.100}{10}=120\left(g\right)\)

c) \(n_{Na2SO4}=\dfrac{0,1.1}{1}=0,1\left(mol\right)\)

⇒ \(m_{Na2SO4}=0,1.142=14,2\left(g\right)\)

\(m_{ddspu}=80+120-5,8=194,2\left(g\right)\)

\(C_{Na2SO4}=\dfrac{14,2.100}{194,2}=7,31\)⁰/₀

 Chúc bạn học tốt

Câu 4 : 

\(n_{Fe}=\dfrac{11,2}{56}=0,2\left(mol\right)\)

Pt : \(Fe+H_2SO_4\rightarrow FeSO_4+H_2|\)

         1         1               1           1

        0,2      0,2

\(n_{H2SO4}=\dfrac{0,2.1}{1}=0,2\left(mol\right)\)

\(m_{H2SO4}=0,2.98=19,6\left(g\right)\)

\(m_{ddH2SO4}=\dfrac{19,6.100}{20}=98\left(g\right)\)

\(V_{ddH2SO4}=\dfrac{98}{1,2}\simeq81,67\left(ml\right)\)

 Chúc bạn học tốt

a) PTHH: \(Mg+2HCl\rightarrow MgCl_2+H_2\uparrow\)

b) Ta có: \(n_{Mg}=\dfrac{3,6}{24}=0,15\left(mol\right)=n_{MgCl_2}\)

\(\Rightarrow m_{MgCl_2}=0,15\cdot95=14,25\left(g\right)\)

c) Theo PTHH: \(\left\{{}\begin{matrix}n_{HCl\left(p.ứ\right)}=0,3\left(mol\right)\\n_{H_2}=0,15\left(mol\right)\end{matrix}\right.\) \(\Rightarrow\left\{{}\begin{matrix}m_{HCl\left(p.ứ\right)}=0,3\cdot36,5=10,95\left(g\right)\\m_{H_2}=0,15\cdot2=0,3\left(g\right)\end{matrix}\right.\)

Mặt khác: \(m_{dd}=m_{Mg}+m_{ddHCl}-m_{H_2}=53,3\left(g\right)\)

\(\Rightarrow m_{ddHCl}=50\left(g\right)\) \(\Rightarrow C\%_{HCl\left(p.ứ\right)}=\dfrac{10,95}{50}\cdot100\%=21,9\%\)  

C là \(BaSO_4\), D là \(HCl\)

\(a,PTHH:BaCl_2+H_2SO_4\rightarrow BaSO_4\downarrow+2HCl\\ n_{BaCl_2}=\dfrac{31,2}{208}=0,15\left(mol\right)\\ \Rightarrow n_{BaSO_4}=0,15\left(mol\right)\\ \Rightarrow m_{BaSO_4}=0,15\cdot233=34,95\left(g\right)\\ b,n_{HCl}=2n_{BaCl_2}=0,3\left(mol\right)\\ \Rightarrow m_{CT_{HCl}}=0,3\cdot36,5=10,95\left(g\right)\\ m_{dd_{HCl}}=31,2+100-34,95=96,25\left(g\right)\\ \Rightarrow C\%_{HCl}=\dfrac{10,95}{96,25}\cdot100\%\approx11,38\%\)

câu 1

cho 2dd trên td vs NaOH dư

có tủa => CuSO4

CuSO4 + 2NaOH => Na2SO4 + Cu(OH)2

ko hiện tượng => Na2SO4

câu 2

nNaOH = 0,75

MgCl2 + 2NaOH => 2NaCl + Mg(OH)2

0,375<---- 0,75--------> 0,75---> 0,375

=> mcr = 0,375. 58 = 21,75 (g)

CM MgCl2 = 0,375/0,5 = 0,75M

SO2+Ba(OH)2->BaSO3+H2O

0,2-----0,2---------0,2--------0,2

n SO2=4,48\22,4=0,2 mol

=>Cm=0,2\0,25=0,8M

=>m BaSO3=0,2.217=43,4g