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 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 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:

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

2Al+3Br₂->2AlBr₃

0,3---0,45----0,3 mol

n Al=\(\dfrac{8,1}{27}\)=0,3 mol

=>m_Br2=0,45.160=72g

=>m _AlBr3=0,3.267=80,1g

\(n_{Al}=\dfrac{8,1}{27}=0,3\left(mol\right)\\ 2Al+3Br_2\rightarrow2AlBr_3\\ n_{AlBr_3}=n_{Al}=0,3\left(mol\right)\\ n_{Br_2}=\dfrac{3}{2}.0,3=0,45\left(mol\right)\\ m_{AlBr_3}=267.0,3=80,1\left(g\right)\\ m_{Br_2}=0,45.160=72\left(g\right)\)

Zn+2HCl->Zncl2+H2

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

n Zn=0,4 mol

VH2=0,4.22,4=8,96l

m ZnCl2=0,4.136=54,4g

2H2+O2-to>2H2O

0,4------0,2----0,4

n O2=0,2 mol

=>pứ hết 

=>m H2O=0,4.18=7,2g

a.b.\(n_{Zn}=\dfrac{26}{65}=0,4mol\)

\(Zn+2HCl\rightarrow ZnCl_2+H_2\)

0,4                      0,4       0,4       ( mol )

\(m_{ZnCl_2}=0,4.136=54,4g\)

\(V_{H_2}=0,4.22,4=8,96l\)

c.\(n_{O_2}=\dfrac{4,48}{22,4}=0,2mol\)

\(2H_2+O_2\rightarrow\left(t^o\right)2H_2O\)

0,4   = 0,2                       ( mol )

0,4        0,2            0,4        ( mol )

\(m_{H_2O}=0,4.18=7,2g\)

nH2 = 0,336/22,4 = 0,015 (mol)

PTHH: 2A + 6HCl -> 2ACl3 + 3H2

nACl3 = nA = 0,015 : 3 . 2 = 0,01 (mol)

M(A) = 0,27/0,01 = 27 (g/mol)

=> A là Al

mAlCl3 = 0,01 . 133,5 = 1,335 (g)

Câu 6.

\(n_{H_2}=\dfrac{0,336}{22,4}=0,015mol\)

\(2A+6HCl\rightarrow2ACl_3+3H_2\)

0,01                                0,015

\(\overline{M_A}=\dfrac{0,27}{0,01}=27đvC\)

\(\Rightarrow A\) là Al nhôm.

\(m_{AlCl_3}=0,01\cdot133,5=1,335g\)