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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

Bài 6:

\(n_{KMnO_4}=\dfrac{31,6}{158}=0,2\left(mol\right)\\ 2KMnO_4+16HCl_{đặc,nóng}\rightarrow2KCl+2MnCl_2+5Cl_2+8H_2O\\ 2Al+3Cl_2\rightarrow\left(t^o\right)2AlCl_3\\ Ta.có:n_{Cl_2}=\dfrac{5}{2}.0,2=0,5\left(mol\right)\\ n_{Al}=\dfrac{10,8}{27}=0,4\left(mol\right)\\ Vì:\dfrac{0,5}{3}< \dfrac{0,8}{2}\Rightarrow Cl_2thiếu,Aldư\)

Bài 7:

\(n_{MnO_2}=\dfrac{52,2}{87}=0,6\left(mol\right)\\ MnO_2+4HBr_{đặc,nóng}\rightarrow MnBr_2+Br_2+2H_2O\\ 2Fe+3Br_2\rightarrow2FeBr_3\\ n_{Br_2}=n_{MnO_2}=0,6\left(mol\right)\\ n_{Fe}=\dfrac{16,8}{56}=0,3\left(mol\right)\\ Vì:\dfrac{0,3}{2}< \dfrac{0,6}{3}\Rightarrow Br_2dư\)

Bài 8:

\(AgNO_3+HCl\rightarrow AgCl\downarrow\left(trắng\right)+HNO_3\\ n_{AgCl}=\dfrac{4,305}{143,5}=0,03\left(mol\right)\\ n_{HCl}=n_{AgCl}=0,03\left(mol\right)\\ V_{ddHCl}=\dfrac{3}{1,15}=\dfrac{60}{23}\left(ml\right)=\dfrac{3}{1150}\left(l\right)\\ C_{MddHCl}=\dfrac{0,03}{\dfrac{3}{1150}}=11,5\left(M\right)\)

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:

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

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

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

Bài 29 :

\(n_{CO2}=\dfrac{4,48}{22,4}=0,2\left(mol\right)\)

Pt : \(CaO+2HCl\rightarrow CaCl_2+H_2O|\)

         1          2               1              1

       0,3       0,6            0,3        

 \(CaCO_3+2HCl\rightarrow CaCl_2+CO_2+H_2O|\)

         1           2             1           1          1

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

a) \(n_{CaCO3}=\dfrac{0,2.1}{1}=0,2\left(mol\right)\)

\(m_{CaCO3}=0,2.100=20\left(g\right)\)

\(m_{CaO}=36,8-20=16,8\left(g\right)\)

⁰/₀CaO = \(\dfrac{16,8.100}{36,8}=45,65\)⁰/₀

⁰/₀CaCO₃ = \(\dfrac{20.100}{36,8}=54,35\)⁰/₀

b) Có : \(m_{CaO}=\dfrac{16,8}{56}=0,3\left(mol\right)\)

\(n_{HCl\left(tổng\right)}=0,6+0,4=1\left(mol\right)\)

\(C_{M_{ddHCl}}=\dfrac{1}{5}=0,2\left(M\right)\)

\(n_{CaCl2\left(tổng\right)}=0,3+0,2=0,5\left(mol\right)\)

\(C_{M_{CaCl2}}=\dfrac{0,5}{5}=0,1\left(M\right)\)

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Bài 28:

\(CaCO_3+2HCl\rightarrow CaCl_2+CO_2+H_2O\\ K_2CO_3+2HCl\rightarrow2KCl+CO_2+H_2O\\ Đặt:n_{CaCO_3}=a\left(mol\right);n_{K_2CO_3}=b\left(mol\right)\left(a,b>0\right)\\ \Rightarrow\left\{{}\begin{matrix}100a+138b=6,76\\a+b=\dfrac{2,64}{44}=0,06\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}a=0,04\\b=0,02\end{matrix}\right.\\ \%m_{CaCO_3}=\dfrac{0,02.100}{6,76}.100\approx29,586\%\\ \Rightarrow\%m_{K_2CO_3}\approx70,414\%\)

câu bao nhiêu vậy

2Al+6HCl->2AlCl₃+3H₂

0,2-----0,6------0,2-----0,3 mol

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

=>V_H2=0,3.22,4=6,72l

=>m _HCl=0,6.36,5=21,9g

=>m _AlCl3=0,2.133,5=26,7g

\(n_{Al}=\dfrac{5,4}{27}=0,2\left(mol\right)\\ 2Al+6HCl\rightarrow\left(t^o\right)2AlCl_3+3H_2\\ n_{AlCl_3}=n_{Al}=0,2\left(mol\right)\\ n_{HCl}=\dfrac{6}{2}.0,2=0,6\left(mol\right)\\ n_{H_2}=\dfrac{3}{2}.0,2=0,3\left(mol\right)\\ a,V_{H_2\left(đktc\right)}=0,3.22,4=6,72\left(l\right)\\ b,m_{HCl}=0,6.36,5=21,9\left(g\right)\\ c,m_{AlCl_3}=133,5.0,2=26,7\left(g\right)\)

Đề bài hỏi gì đó em?

\(n_{H_2}=\dfrac{4,48}{22,4}=0,2\left(mol\right)\\ 2Na+2H_2O\rightarrow2NaOH+H_2\\ Ca+2H_2O\rightarrow Ca\left(OH\right)_2+H_2\\ Đặt:n_{Na}=a\left(mol\right);n_{Ca}=b\left(mol\right)\left(a,b>0\right)\\ \Rightarrow\left\{{}\begin{matrix}23a+40b=8,3\\0,5a+b=0,2\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}a=0,1\\b=0,15\end{matrix}\right.\\ b,\Rightarrow\%m_{Ca}=\dfrac{0,15.40}{8,3}.100\approx72,289\%\\ \Rightarrow\%m_{Na}\approx27,711\%\\ b,n_{NaOH}=a=0,1\left(mol\right)\\ n_{Ca\left(OH\right)_2}=b=0,15\left(mol\right)\\ m_{bazo}=m_{NaOH}+m_{Ca\left(OH\right)_2}=40.0,1+74.0,15=15,1\left(g\right)\)

2Na+2H₂O->2NaOH+H₂

x------------------------------0,5x

Ca+2H₂O->Ca(OH)₂+H₂

y-------------------------------y 

Ta có :

\(\left\{{}\begin{matrix}23x+40y=8,3\\0,5x+y=0,2\end{matrix}\right.=>\left\{{}\begin{matrix}x=0,1\\y=0,15\end{matrix}\right.\)

=>%mNa=\(\dfrac{0,1.23}{8,3}.100=27,71\%\)

=>%mCa=72,29%

b)m bazo=0,1.40+0,15.74=15,1g

\(Ca+2H_2O\rightarrow Ca\left(OH\right)_2+H_2\)

    Ca + 2H2O → Ca(OH)2 + H2 ↑ nhé

Ca+2H₂O->Ca(OH)₂+H₂