Câu 2 : 

\(n_{Cu}=\dfrac{22,4}{64}=0,35\left(mol\right)\)

Pt : \(Cu+Cl_2\underrightarrow{t^o}CuCl_2|\)

         1         1            1

        0,35    0,35       0,35

 \(n_{CuCl2}=\dfrac{0,35.1}{1}=0,35\left(mol\right)\)

⇒ \(m_{CuCl2}=0,35.135=47,25\left(g\right)\)

\(n_{Cl2}=\dfrac{0,35.1}{1}=0,35\left(mol\right)\)

\(V_{Cl2\left(dtkc\right)}=0,35.22,4=7,84\left(l\right)\)

 Chúc bạn học tốt

\(Câu4\\ n_{Cl_2}=\dfrac{3,36}{22,4}=0,15\left(mol\right)\\ 2Al+3Cl_2\rightarrow\left(t^o\right)2AlCl_3\\ n_{Al}=n_{AlCl_3}=\dfrac{2}{3}.0,15=0,1\left(mol\right)\\ \Rightarrow m=m_{Al}=0,1.27=2,7\left(g\right)\\ m_{AlCl_3}=133,5.0,1=13,35\left(g\right)\)

Câu 3:

\(n_{Br_2}=\dfrac{10,4}{160}=0,065\left(mol\right)\\ 2Na+Br_2\rightarrow2NaBr\\ n_{Na}=n_{NaBr}=0,065.2=0,13\left(mol\right)\\ a=m_{Na}=0,13.23=2,99\left(g\right)\\ m_{NaBr}=0,13.103=13,39\left(g\right)\)

Câu 2:

\(n_{MgBr_2}=\dfrac{14,72}{184}=0,08\left(mol\right)\\ Mg+Br_2\rightarrow MgBr_2\\ n_{Mg}=n_{Br_2}=n_{MgBr_2}=0,08\left(mol\right)\\ a=m_{Mg}=24.0,08=1,92\left(g\right)\\ m_{Br_2}=160.0,08=12,8\left(g\right)\)

Câu 1:

\(n_{AlBr_3}=\dfrac{106,8}{267}=0,4\left(mol\right)\\ 2Al+3Br_2\rightarrow2AlBr_3\\ n_{Al}=n_{AlBr_3}=0,4\left(mol\right)\\ n_{Br_2}=\dfrac{3}{2}.0,4=0,6\left(mol\right)\\ a=m_{Al}=0,4.27=10,8\left(g\right)\\ m_{Br_2}=160.0,6=96\left(g\right)\)

2Al+3Br₂-to>2AlBr₃

0,4-----0,6-------0,4 mol

n AlBr3=\(\dfrac{106,8}{267}\)=0,4 mol

=>a=mAl=0,4.27=10,8g

=>mBr₂=0,6.160=96g

Câu 2

Mg+Br₂->MgBr₂

0,08--0,08---0,08 mol

n _MgBr2=\(\dfrac{14,72}{184}\)=0,08 mol

=>a=mMg=0,08.24=1,92g

=>m _Br2=0,08.160=12,8g

Câu 1:

\(2Na+Br_2\rightarrow2NaBr\\ n_{NaBr}=\dfrac{61,8}{103}=0,6\left(mol\right)\\ n_{Na}=n_{NaBr}=0,6\left(mol\right)\\ n_{Br_2}=\dfrac{0,6}{2}=0,3\left(mol\right)\\ \Rightarrow a=m_{Na}=0,6.23=13,8\left(g\right)\\ m_{Br_2}=0,3.160=48\left(g\right)\\ m_{ddBr_2}=\dfrac{48}{5\%}=960\left(g\right)\)

Câu 2:

\(2Fe+3Cl_2\rightarrow\left(t^o\right)2FeCl_3\\ n_{FeCl_3}=\dfrac{40,625}{162,5}=0,25\left(mol\right)\\ n_{Fe}=n_{FeCl_3}=0,25\left(mol\right)\\ \Rightarrow m=m_{Fe}=0,25.56=14\left(g\right)\\ n_{Cl_2}=\dfrac{3}{2}.0,25=0,375\left(mol\right)\\ V_{Cl_2\left(đktc\right)}=0,375.22,4=8,4\left(l\right)\)

2Fe+3Cl₂-to>2FeCl₃

_{0,25------0,375------0,25 mol}

n _FeCl3=\(\dfrac{40,625}{162,5}\)=0,25 mol

=>m _Fe=0,25.56=14g

=>V _Cl2=0,375.22,4=8,4g

Câu 1:

\(Mg+Br_2\rightarrow MgBr_2\\ n_{Br_2}=\dfrac{11,2}{160}=0,07\left(mol\right)=n_{Mg}=n_{MgBr_2}\\ a=m_{Mg}=0,07.24=1,68\left(g\right)\\ m_{MgBr_2}=184.0,07=12,88\left(g\right)\)

Mg+Br2->MgBr2

0,07--0,07----0,07

n Br2=\(\dfrac{11,2}{160}\)=0,07 mol

=>m Mg=0,07.24=1,68g

=>m MgBr2=0,07.184=12,88g

Câu 2:

\(n_{Al}=\dfrac{14,58}{27}=0,54\left(mol\right)\\ 2Al+3Cl_2\rightarrow\left(t^o\right)2AlCl_3\\ n_{AlCl_3}=n_{Al}=0,54\left(mol\right)\\ n_{Cl_2}=\dfrac{3}{2}.0,54=0,81\left(mol\right)\\ V_{Cl_2\left(đktc\right)}=0,81.22,4=18,144\left(l\right)\\ m_{AlCl_3}=0,54.133,5=72,09\left(g\right)\)

\(a,PTHH:R+2AgNO_3\to R(NO_3)_2+2Ag\\ \Rightarrow n_{R}=n_{R(NO_3)_2}\\ \Rightarrow \dfrac{2,8}{M_R}=\dfrac{9}{M_R+124}\\ \Rightarrow M_R=56(g/mol)\)

Vậy R là sắt (Fe)

\(b,n_{R}=\dfrac{2,8}{56}=0,05(mol)\\ \Rightarrow n_{AgNO_3}=0,1(mol)\\ \Rightarrow m_{dd_{AgNO_3}}=\dfrac{0,1.170}{5\%}=340(g)\\ c,n_{Fe(NO_3)_2}=n_{Fe}=0,05(mol);n_{Ag}=0,1(mol)\\ \Rightarrow C\%_{Fe(NO_3)_2}=\dfrac{0,05.180}{2,8+340-0,1.108}.100\%=2,71\%\)

a) \(n_{Mg}=\dfrac{2,4}{24}=0,1\left(mol\right)\)

PTHH: Mg + H₂SO₄ --> MgSO₄ + H₂

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

=> \(m_{dd.H_2SO_4}=\dfrac{0,1.98}{4,9\%}=200\left(g\right)\)

b) m_{dd sau pư} = 2,4 + 200 - 0,1.2 = 202,2 (g)

m_MgSO4 = 0,1.120 = 12 (g)

\(C\%_{MgSO_4}=\dfrac{12}{202,2}.100\%=5,9\%\)

c) 

\(n_{CuO}=\dfrac{20}{80}=0,25\left(mol\right)\)

PTHH: CuO + H₂ --t^o--> Cu + H₂O

Xét tỉ lệ: \(\dfrac{0,25}{1}>\dfrac{0,1}{1}\) => Hiệu suất tính theo H₂

\(n_{Cu}=\dfrac{3,2}{64}=0,05\left(mol\right)\)

PTHH: CuO + H₂ --t^o--> Cu + H₂O

                      0,05<-----0,05

=> \(H=\dfrac{0,05}{0,1}.100\%=50\%\)

a, \(n_{Mg}=\dfrac{2,4}{24}=0,1\left(mol\right)\)

PTHH: Mg + H₂SO₄ ---> MgSO₄ + H₂

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

\(m_{dd\left(H_2SO_4\right)}=\dfrac{0,1.98}{4,9\%}=200\left(g\right)\)

b, \(m_{dd\left(sau.pư\right)}=2,4+200-0,2.2=202,2\left(g\right)\)

\(\rightarrow C\%_{MgSO_4}=\dfrac{0,1.120}{202,2}.100\%=5,93\%\)

c, \(n_{CuO}=\dfrac{20}{80}=0,25\left(mol\right)\)

PTHH: CuO + H₂ --t^o--> Cu + H₂O

LTL: 0,25 > 0,1 => CuO dư

\(n_{Cu}=\dfrac{3,2}{64}=0,05\left(mol\right)\)

Theo pt: \(n_{H_2}=n_{Cu}=0,05\left(mol\right)\)

=> \(H=\dfrac{0,05}{0,1}.100\%=50\%\)

\(n_{Al}=\dfrac{5,4}{27}=0,2mol\)

a)\(2Al+6HCl\rightarrow2AlCl_3+3H_2\)

    0,2     0,6                         0,3

\(C_M=\dfrac{0,6}{0,4}=1,5M\)

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

\(CuO+H_2\rightarrow Cu+H_2O\)

0,4        0,3     0,3

Sau phản ứng CuO dư và dư \(\left(0,4-0,3\right)\cdot80=8g\)

\(m_{rắn}=m_{Cu}=0,3\cdot64=19,2g\)

Em đăng sang môn Hoá nha

\(2Al+6HCl->2AlCl_3+3H_2\\ Fe+2HCl->FeCl_2+H_2\\ n_{Al}=a;n_{Fe}=b\\ 27a+56b=8,3\\ 1,5a+b=\dfrac{5,6}{22,4}=0,25\\ a=b=0,1\\ m_{Al}=27\cdot0,1=2,7g\\ m_{Fe}=8,3-2,7=5,6g\\ a=\dfrac{3a+2b}{500}\cdot36,5=3,65\%\\ m_{ddsau}=508,3-0,25\cdot2=507,8g\\ C\%_{AlCl_3}=\dfrac{133,5a}{507,8}=2,63\%\\ C\%_{FeCl_2}=\dfrac{127b}{507,8}=2,50\%\)