Câu 1 : 

\(n_{Al}=\dfrac{2.7}{27}=0.1\left(mol\right)\)

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

\(0.1......................0.1.........0.15\)

\(m_{dd}=2.7+200-0.15\cdot2=202.4\left(g\right)\)

\(C\%_{AlCl_3}=\dfrac{0.1\cdot133.5}{202.4}\cdot100\%=6.59\%\)

\(n_{ZnO}=\dfrac{8,1}{81}=0,1\left(mol\right)\\ n_{H_2SO_4}=\dfrac{19,6\%.200}{98}=0,4\left(mol\right)\\a, ZnO+H_2SO_4\rightarrow ZnSO_4+H_2O\\ b,Vì:\dfrac{0,1}{1}< \dfrac{0,4}{1}\\ \Rightarrow H_2SO_4dư\\ n_{H_2SO_4\left(p.ứ\right)}=n_{ZnSO_4}=n_{ZnO}=0,1\left(mol\right)\\ n_{H_2SO_4\left(dư\right)}=0,4-0,1=0,3\left(mol\right)\\ \Rightarrow m_{H_2SO_4\left(dư\right)}=98.0,3=29,4\left(g\right)\\ c,n_{ZnSO_4}=0,1.161=16,1\left(g\right)\\ m_{ddsau}=m_{ZnO}+m_{ddH_2SO_4}=8,1+200=208,1\left(g\right)\\ \Rightarrow C\%_{ddH_2SO_4\left(dư\right)}=\dfrac{29,4}{208,1}.100\approx14,128\%\\ C\%_{ddZnSO_4}=\dfrac{16,1}{208,1}.100\approx7,737\%\)

ZnO+H₂SO₄->ZnSO₄+H₂O

0,1-----0,1-------0,1-------0,1 mol

n _ZnO=\(\dfrac{8,1}{81}\)=0,1 mol

m _H2SO4 =39,2g =>n _H2SO4=\(\dfrac{39,2}{98}\)=0,4 mol

=>H2SO4 , dư 0,3 mol

=>m _H2SO4=0,3.98=29,4g

=>C%_H2SO4 dư=\(\dfrac{29,4}{200+0,1.18}\).100=14,568%

=>C% _ZnSO4=\(\dfrac{0,1.161}{200+0,1.18}.100=7,9781\%\)

\(n_{CO_2}=\dfrac{V}{22,4}=\dfrac{4,48}{22,4}=0,2\left(mol\right)\\ n_{NaOH}=\dfrac{m}{M}=\dfrac{m_{dd}.C\%}{M}=\dfrac{200.16\%}{40}=0,8\left(mol\right)\)

Có: \(\dfrac{n_{NaOH}}{n_{CO_2}}=4\)

=> Phản ứng tạo muối Na₂CO₃

PT:

\(CO_2+2NaOH\rightarrow Na_2CO_3+H_2O\)

0,2.       0,4            0,2

=> dd sau phản ứng có những chất tan là:

\(\left\{{}\begin{matrix}Na_2CO_3:0,2\left(mol\right)\\NaOH:0,4\left(mol\right)\end{matrix}\right.\)

m_dd spu=0,2.44+200=208,8(g)

\(\%m_{NaOH}=\dfrac{0,4.40}{208,8}.100\%=7,66\%\\\%m_{Na_2CO_3}=\dfrac{0,2.106}{208,8}.100\%=10,15\% \)

Câu 1:

nAl= 0,1(mol)

PTHH: 2Al + 6 HCl -> 2 AlCl3 + 3 H2 

nAlCl3= nAl=0,1(mol)

-> mAlCl3= 133,5 x 0,1= 13,35(g)

mddAlCl3= mAl + mddHCl - mH2 = 2,7 + 200 - 3/2 x 0,1 x 2= 202,4(g)

C%ddAlCl3= (13,35/202,4).100= 6,596%

bạn ơi, có nồng độ của dd HCl không? vì nếu làm ntn thì câu 1 đề bài không chặt chẽ

\(a.n_{H_2SO_4}=\dfrac{294.10\%}{98}=0,3\left(mol\right)\\ n_{MgO}=\dfrac{8}{40}=0,2\left(mol\right)\\ MgO+H_2SO_4\rightarrow MgSO_4+H_2O\\ Vì:\dfrac{0,3}{1}>\dfrac{0,2}{1}\\ \rightarrow H_2SO_4dư\\ n_{MgSO_4}=n_{H_2SO_4\left(p.ứ\right)}=n_{MgO}=0,2\left(mol\right)\\ n_{H_2SO_4\left(p.ứ\right)}=0,3-0,2=0,1\left(mol\right)\\ m_{ddsau}=8+294=302\left(g\right)\\ b.C\%_{ddH_2SO_4\left(Dư\right)}=\dfrac{0,1.98}{302}.100\approx3,245\%\\ C\%_{ddMgSO_4}=\dfrac{0,2.120}{302}.100\approx7,947\%\)

Ta có:

 n MgO = 0,2 ( mol )

m H2SO4 = 294 . 10% = 29,4 ( g )

=> n H2SO4 = 0,3 ( mol )

PTHH

    MgO + H2SO4 ====> MgSO4 + H2O

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

 theo pthh: n H2SO4 phản ứng = n Mg = 0,2 ( mol )

=> n H2SO4 dư = 0,1 ( mol )

BTKL:

 m dd sau phản ứng = 8 + 294 = 302 ( g )

    => %m H2SO4 dư = 3,25 %

          %m MgSO4 = 7,95%

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

\(m_{ct}=\dfrac{294.10}{100}=29,4\left(g\right)\)

\(n_{H2SO4}=\dfrac{29,4}{98}=0,3\left(mol\right)\)

a) Pt : \(MgO+H_2SO_4\rightarrow MgSO_4+H_2O|\)

         1             1                1             1

        0,2           0,3            0,2

b)Lập tỉ số so sánh : \(\dfrac{0,2}{1}< \dfrac{0,3}{1}\)

                 ⇒ MgO phản ứng hết , H₂SO₄ dư

                ⇒ Tính toán dựa vào số mol của MgO

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

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

\(n_{H2SO4\left(dư\right)}=0,3-0,2=0,1\left(mol\right)\)

⇒ \(m_{H2SO4\left(dư\right)}=0,1.98=9,8\left(g\right)\)

Sau phản ứng :

\(m_{dd}=8+294=302\left(g\right)\)

\(C_{MgSO4}=\dfrac{12.100}{302}=3,97\)⁰/₀

\(C_{H2SO4\left(dư\right)}=\dfrac{9.8.100}{302}=3,25\)⁰/₀

 Chúc bạn học tốt

a) \(n_{FeCl_3}=\dfrac{16,25}{162,5}=0,1\left(mol\right)\)

PTHH: Fe₂O₃ + 6HCl --> 2FeCl₃ + 3H₂O

            0,05<----0,3<-----0,1

=> \(m_{Fe_2O_3}=0,05.160=8\left(g\right)\)

b) 

\(m_{HCl\left(bd\right)}=91,25.16\%=14,6\left(g\right)\)

m_{dd sau pư} = 8 + 91,25 = 99,25 (g)

\(\left\{{}\begin{matrix}C\%\left(FeCl_3\right)=\dfrac{16,25}{99,25}.100\%=16,373\%\\C\%\left(HCldư\right)=\dfrac{14,6-0,3.36,5}{99,25}.100\%=3,678\%\end{matrix}\right.\)

\(n_{Fe_2O_3}=\dfrac{16}{160}=0,1mol\)

\(Fe_2O_3+3H_2SO_4\rightarrow Fe_2\left(SO_4\right)_3+3H_2O\)

0,1             0,3                 0,1              0,3

\(m_{H_2SO_4}=0,3\cdot98=29,4\left(g\right)\)\(\Rightarrow m_{ddH_2SO_4}=\dfrac{29,4}{25}\cdot100=117,6\left(g\right)\)

\(m_{H_2O}=0,3\cdot18=5,4\left(g\right)\)

\(m_{Fe_2\left(SO_4\right)_3}=0,1\cdot400=40\left(g\right)\)

\(m_{ddsau}=16+117,6-5,4=128,2\left(g\right)\)

\(C\%=\dfrac{40}{128,2}\cdot100\%=31,2\%\)

Ta có: \(n_{Fe_2O_3}=\dfrac{16}{160}=0,1\left(mol\right)\)

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

Theo PT: \(n_{H_2SO_4}=3.n_{H_2SO_4}=3.0,1=0,3\left(mol\right)\)

=> \(m_{H_2SO_4}=0,3.98=29,4\left(g\right)\)

Ta có: \(C_{\%_{H_2SO_4}}=\dfrac{29,4}{m_{dd_{H_2SO_4}}}.100\%=25\%\)

=> \(m_{dd_{H_2SO_4}}=117,6\left(g\right)\)

=> \(m_{dd_{Fe_2\left(SO_4\right)_3}}=117,6+16=133,6\left(g\right)\)

Theo PT: \(n_{Fe_2\left(SO_4\right)_3}=n_{Fe_2O_3}=0,1\left(mol\right)\)

=> \(m_{Fe_2\left(SO_4\right)_3}=0,1.400=40\left(g\right)\)

=> \(C_{\%_{Fe_2\left(SO_4\right)_3}}\dfrac{40}{133,6}.100\%=29,94\%\)