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

4Al+3O₂-to>2Al₂O₃

0,2---0,15-------0,1 mol

m _Al2O3=0,1.102=10,2g

V_kk=0,15.22,4.5=16,8l

\(a,4Al+3O_2\rightarrow\left(t^o\right)2Al_2O_3\\ b,n_{Al}=\dfrac{5,4}{27}=0,2\left(mol\right)\\ n_{Al_2O_3}=\dfrac{2}{4}.0,2=0,1\left(mol\right)\\ \Rightarrow m_{Al_2O_3}=102.0,1=10,2\left(g\right)\\ c,n_{O_2}=\dfrac{3}{4}.0,2=0,15\left(mol\right)\\ V_{H_2\left(đktc\right)}=0,15.22,4=3,36\left(l\right)\\ V_{kk\left(đktc\right)}=5.3,36=16,8\left(l\right)\)

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

\(n_{Al}=\dfrac{5.4}{27}=0.2\left(mol\right)\)

\(n_{HCl}=\dfrac{10.95}{36.5}=0.3\left(mol\right)\)

Vì \(\dfrac{0.2}{2}>\dfrac{0.3}{6}\) nên Al dư và dư 0,05 mol

b: \(n_{AlCl_3}=0.9\left(mol\right)\)

\(m_{AlCl_3}=0.9\cdot136.5=122.85\left(g\right)\)

\(a,n_{Fe}=\dfrac{5,6}{56}=0,1\left(mol\right)\\PTHH:Fe+2HCl\rightarrow FeCl_2+H_2\\ Tỉ.lệ.p.ứ.hh=1:2:1:1\\ b,n_{H_2}=n_{Fe}=0,1\left(mol\right)\\ V_{H_2\left(đktc\right)}=0,1.22,4=2,24\left(l\right)\\ c,n_{HCl}=2.0,1=0,2\left(mol\right)\)

a. \(n_{Fe}=\dfrac{5,6}{56}=0,1\left(mol\right)\)

PTHH : Fe + 2HCl -> FeCl₂ + H₂

Tỉ lệ : 1 : 2 : 1 :1

Số mol : 0,1    0,2       0,1        0,1 

b. \(V_{H_2}=0,1.22,4=2,24\left(l\right)\)

Đề bài là 8,96 lít hay 9,96 lít em hầy?

Chắc 8,96 chứ 9,96/22,4 = R

Gọi số mol của Al₂O₃, Fe₂O₃ là 2a, 3a (mol)

PTHH: Al₂O₃ + 3H₂SO₄ --> Al₂(SO₄)₃ + 3H₂O

              2a---->6a

            Fe₂O₃ + 3H₂SO₄ --> Fe₂(SO₄)₃ + 3H₂O

               3a----->9a

=> 6a + 9a = 0,3

=> a = 0,02 

=> \(\left\{{}\begin{matrix}n_{Al_2O_3}=0,04\left(mol\right)\\n_{Fe_2O_3}=0,06\left(mol\right)\end{matrix}\right.\)

=> \(\left\{{}\begin{matrix}\%m_{Al_2O_3}=\dfrac{0,04.102}{0,04.102+0,06.160}.100\%=29,825\%\\\%m_{Fe_2O_3}=\dfrac{0,06.160}{0,04.102+0,06.160}.100\%=70,175\%\end{matrix}\right.\)

\(n_{Mg}=\dfrac{2,4}{24}=0,1\left(mol\right);n_{HCl}=\dfrac{3,65}{36,5}=0,1\left(mol\right)\\ a,Mg+2HCl\rightarrow MgCl_2+H_2\\ Vì:\dfrac{0,1}{2}< \dfrac{0,1}{1}\\ \Rightarrow Mgdư\\ \Rightarrow n_{Mg\left(p.ứ\right)}=n_{MgCl_2}=n_{H_2}=\dfrac{0,1}{2}=0,05\left(mol\right)\\ n_{Mg\left(dư\right)}=0,1-0,05=0,05\left(mol\right)\\ \Rightarrow m_{Mg\left(dư\right)}=0,05.24=1,2\left(g\right)\\ b,m_{MgCl_2}=95.0,05=4,75\left(g\right)\\ c,V_{H_2\left(đktc\right)}=0,05.22,4=1,12\left(l\right)\)

a) \(n_{CO_2}=\dfrac{0,672}{22,4}=0,03\left(mol\right)\)

Gọi công thức chung 2 muối cacbonat của A, B là XCO₃

PTHH: XCO₃ + 2HCl --> XCl₂ + CO₂ + H₂O

=> \(M_{XCO_3}=\dfrac{2,84}{0,03}=94,667\left(g/mol\right)\)

=> M_X = 34,667 g/mol

Mà A, B kế tiếp nhau trong nhóm IIA

=> A,B là Mg, Ca

b)

Gọi số mol MgCO₃, CaCO₃ là a, b (mol)

=> 84a + 100b = 2,84

PTHH: MgCO₃ + 2HCl --> MgCl₂ + CO₂ + H₂O

               a------------------>a------->a

             CaCO₃ + 2HCl --> CaCl₂ + CO₂ + H₂O

               b-------------------->b------->b

=> a + b = 0,03 

=> a = 0,01; b = 0,02

=> \(\left\{{}\begin{matrix}m_{MgCl_2}=0,01.95=0,95\left(g\right)\\m_{CaCl_2}=0,02.111=2,22\left(g\right)\end{matrix}\right.\)

\(n_{CO_2}=\dfrac{0,672}{22,4}=0,03\left(mol\right)\Rightarrow n_{hh}=n_{CO^{2-}_3}=0,03\left(mol\right)\\ \Rightarrow m_{A,B}=2,84-0,03.60=1,04\left(g\right)\\24< M_{A,B}=\dfrac{1,04}{0,03}\approx34,667< 40\\ \Rightarrow A,B:Magie\left(Mg\right),Canxi\left(Ca\right)\\ b,Đặt:n_{MgCO_3}=a\left(mol\right);n_{CaCO_3}=b\left(mol\right)\left(a,b>0\right)\\ MgCO_3+2HCl\rightarrow MgCl_2+CO_2+H_2O\\ CaCO_3+2HCl\rightarrow CaCl_2+CO_2+H_2O\\ \Rightarrow\left\{{}\begin{matrix}84a+100b=2,84\\a+b=0,03\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}a=0,01\\b=0,02\end{matrix}\right.\\ \Rightarrow\%m_{CaCO_3}=\dfrac{0,02.100}{2,84}.100\approx70,423\%\\ \Rightarrow\%m_{MgCO_3}\approx29,577\%\)

 Gọi số mol Fe phản ứng là a (mol)

PTHH: Fe + CuSO₄ --> FeSO₄ + Cu

            a------------------->a----->a

=> 50 - 56a + 64a = 51

=> a = 0,125 (mol)

=> \(n_{FeSO_4}=0,125\left(mol\right)\)

=> \(m_{FeSO_4}=0,125.152=19\left(g\right)\)

\(Đặt:n_{Fe\left(pứ\right)}=x\left(mol\right)\\ Fe+CuSO_4\rightarrow FeSO_4+Cu\\m_{tăng}=m_{Cu}-m_{Fe\left(pứ\right)}=51-50\\ \Leftrightarrow 64x-56x=1\\ \Rightarrow x=0,125\left(mol\right)\\ n_{FeSO_4}=n_{Fe}=0,125\left(mol\right)\\ \Rightarrow m_{FeSO_4}=0,125.152=19\left(g\right)\)

trạng thái ko màu , ko mùi , nhẹ hơn kk , dạng khí 

TK

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