a) Khí sinh ra là \(H_2\)

\(n_{H_2}=\dfrac{4,48}{22,4}=0,2\left(mol\right)\)

PTHH:

\(Mg+H_2SO_4->MgSO_4+H_2\)

0,2      0,2                0,2    <--  0,2  (mol)

\(MgO+H_2SO_4->MgSO_4+H_2O\)

0,08  -->    0,08       0,08            0,08 (mol)

\(m_{Mg}=0,2\cdot24=4,8\left(g\right)\)

--> \(m_{MgO}=m_{hh}-m_{Mg}=8-4,8=3,2\left(g\right)\\ n_{MgO}=\dfrac{3,2}{40}=0,08\left(mol\right)\)

--> \(\%m_{Mg}=\dfrac{m_{Mg}}{m_{hh}}\cdot100\%=\dfrac{4,8}{8}\cdot100\%=60\%\)

--> \(\%m_{MgO}=\dfrac{m_{MgO}}{m_{hh}}\cdot100\%=\dfrac{3,2}{8}\cdot100\%=40\%\)

b) Đổi 200ml = 0,2l

\(n_{H_2SO_4}=0,2+0,08=0,28\left(mol\right)\)

\(C_{M\left(H_2SO_4\right)}=\dfrac{0,28}{0,2}=1,4\left(M\right)\)

\(m_{Al\left(NO_3\right)_3}=\dfrac{300.4,26}{100}=12,78\left(g\right)\)

\(n_{Al\left(NO_3\right)_3}=\dfrac{12,78}{213}=0,06\left(mol\right)\)

\(m_{NaOH}=\dfrac{200.4,2}{100}=8,4\left(g\right)\)

\(n_{NaOH}=\dfrac{8,4}{40}=0,21\left(mol\right)\)

\(Al\left(NO_3\right)_3+3NaOH\rightarrow NaNO_3+Al\left(OH\right)_3\)

0,06               0,18           0,06             0,06

LTL: \(\dfrac{0,06}{1}< \dfrac{0,21}{3}\) => NaOH dư

\(n_{NaOH\left(dư\right)}=0,21-0,18=0,03\left(mol\right)\)

dd X là \(NaNO_3\left(sinh.ra\right)\)  và \(NaOH\left(dư\right)\)

kt Y là \(Al\left(OH\right)_3\)       

\(m_{kt}=m_{Al\left(OH\right)_3}=0,06.78=4,68\left(g\right)\)

\(m_{dd}=300+200-4,68=495,32\left(g\right)\)

\(m_{NaOH\left(dư\right)}=0,03.40=1,2\left(g\right)\)

\(m_{NaNO_3}=0,06.85=5,1\left(g\right)\)

\(C\%_{ddNaOH}=\dfrac{1,2.100}{495,32}=0,24\%\)

\(C\%_{ddNaNO_3}=\dfrac{5,1.100}{495,32}=1,03\%\)

Giả sử có 1 mol A

PTHH: \(2A+nH_2SO_4\rightarrow A_2\left(SO_4\right)_n+nH_2\)

              1------------------------------>0,5n

            \(3A+4mHNO_3\rightarrow3A\left(NO_3\right)_m+mNO+2mH_2O\)

               1---------------------------------->\(\dfrac{m}{3}\)

=> \(\dfrac{n_{NO}}{n_{H_2}}=\dfrac{\dfrac{m}{3}}{0,5n}=1\Rightarrow\dfrac{m}{n}=\dfrac{3}{2}\)

Vậy m = 3; n = 2

Muối X có dạng ASO₄, muối Y có dạng A(NO₃)₃

Ta có: \(\dfrac{NTK_A+96}{NTK_A+186}=\dfrac{152}{242}\) => NTK_A = 56 (đvC)

=> A là Fe

X là FeSO₄, Y là Fe(NO₃)₃

Gọi số mol CuO, Fe₂O₃ là a, b (mol)

=> 80a + 160b = m (1)

PTHH: \(CuO+CO\underrightarrow{t^o}Cu+CO_2\)

               a----------->a---->a

            \(Fe_2O_3+3CO\underrightarrow{t^o}2Fe+3CO_2\)

                 b------------->2b---->3b

=> 64a + 112b = n (2)

PTHH: \(Ca\left(OH\right)_2+CO_2\rightarrow CaCO_3+H_2O\)

                             (a+3b)--->(a+3b)

=> 100a + 300b = p (3)

(1)(2)(3) => m = n + 0,16p

a) \(n_{Al_2\left(SO_4\right)_3}=0,5.0,2=0,1\left(mol\right)\)

\(n_{Ba\left(OH\right)_2}=0,8.0,4=0,32\left(mol\right)\)

PTHH: \(Al_2\left(SO_4\right)_3+3Ba\left(OH\right)_2\rightarrow3BaSO_4\downarrow+2Al\left(OH\right)_3\downarrow\)

                   0,1----------->0,3------------>0,3--------->0,2 

             \(2Al\left(OH\right)_3+Ba\left(OH\right)_2\rightarrow Ba\left(AlO_2\right)_2+4H_2O\)

                   0,04<-------0,02

Kết tủa gồm \(\left\{{}\begin{matrix}BaSO_4:0,3\left(mol\right)\\Al\left(OH\right)_3:0,16\left(mol\right)\end{matrix}\right.\)

=> m_kt = 0,3.233 + 0,16.78 = 82,38 (g)

c)

\(\left\{{}\begin{matrix}n_{AlCl_3}=1.0,1=0,1\left(mol\right)\\n_{KOH}=1.0,5=0,5\left(mol\right)\end{matrix}\right.\)

PTHH: \(3KOH+AlCl_3\rightarrow Al\left(OH\right)_3\downarrow+3KCl\)

                 0,3<-----0,1-------->0,1

            \(KOH+Al\left(OH\right)_3\rightarrow KAlO_2+2H_2O\)

                0,1<------0,1

=> Không có kết tủa

d)

\(\left\{{}\begin{matrix}n_{NaAlO_2}=0,25.2=0,5\left(mol\right)\\n_{H_2SO_4}=2.0,375=0,75\left(mol\right)\end{matrix}\right.\)

PTHH: \(2NaAlO_2+H_2SO_4+2H_2O\rightarrow Na_2SO_4+2Al\left(OH\right)_3\downarrow\)

                  0,5------->0,25------------------------------->0,5

             \(2Al\left(OH\right)_3+3H_2SO_4\rightarrow Al_2\left(SO_4\right)_3+6H_2O\)

                    \(\dfrac{1}{3}\)<--------0,5

=> \(m_{Al\left(OH\right)_3}=\left(0,5-\dfrac{1}{3}\right).78=13\left(g\right)\)

e) \(\left\{{}\begin{matrix}n_{H_2SO_4}=0,6.0,5=0,3\left(mol\right)\\n_{K_2ZnO_2}=1.0,5=0,5\left(mol\right)\end{matrix}\right.\)

PTHH: \(K_2ZnO_2+H_2SO_4\rightarrow K_2SO_4+Zn\left(OH\right)_2\downarrow\)

                                    0,3-------------------->0,3

=> \(m_{Zn\left(OH\right)_2}=0,3.99=29,7\left(g\right)\)

Quy đổi hỗn hợp ban đầu thành kim loai X, hóa trị n

\(n_X=\dfrac{20}{M_X}\left(mol\right)\)

PTHH: \(4X+nO_2\underrightarrow{t^o}2X_2O_n\)

           \(\dfrac{20}{M_X}\)-------->\(\dfrac{10}{M_X}\)

            \(X_2O_n+2nHCl\rightarrow2XCl_n+nH_2O\)

              \(\dfrac{10}{M_X}\)----->\(\dfrac{20n}{M_X}\)----->\(\dfrac{20}{M_X}\)

Ta có: \(m_{MCl_n}=\dfrac{20}{M_X}\left(M_X+35,5n\right)=27,1\left(g\right)\)

=> \(\)\(M_X=100n\left(g/mol\right)\)

Ta có: \(m_{X_2O_n}=\dfrac{10}{M_X}\left(2.M_X+16n\right)=21,6\left(g\right)\)

Ta có: \(n_{HCl}=0,2\left(mol\right)\) => \(C_{M\left(HCl\right)}=\dfrac{0,2}{0,1}=2M\)

bn check lại đề bn nhé, 20(g) kim loại mà thu được 2,1(g) muối thì nghe vô lí đấy :)

a) \(n_{CuSO_4}=\dfrac{80.40\%}{160}=0,2\left(mol\right)\)

\(n_{NaOH}=\dfrac{100.20\%}{40}=0,5\left(mol\right)\)

PTHH: \(CuSO_4+2NaOH\rightarrow Cu\left(OH\right)_2\downarrow+Na_2SO_4\)

Xét tỉ lệ: \(\dfrac{0,2}{1}< \dfrac{0,5}{2}\) => CuSO₄ hết, NaOH dư

PTHH: \(CuSO_4+2NaOH\rightarrow Cu\left(OH\right)_2\downarrow+Na_2SO_4\)

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

=> m_NaOH(dư) = (0,5 - 0,4).40 = 4 (g)

b) m_{dd sau pư} = 80 + 100 - 0,2.98 = 160,4 (g)

\(\left\{{}\begin{matrix}C\%_{NaOH\left(dư\right)}=\dfrac{4}{160,4}.100\%=2,5\%\\C\%_{Na_2SO_4}=\dfrac{0,2.142}{160,4}.100\%=17,7\%\end{matrix}\right.\)

a) \(n_{CuSO_4}=\dfrac{80.40\%}{160}=0,2\left(mol\right);n_{NaOH}=\dfrac{100.20\%}{40}=0,5\left(mol\right)\)

PTHH:     \(CuSO_4+2NaOH\rightarrow Cu\left(OH\right)_2\downarrow+Na_2SO_4\)

ban đầu    0,2            0,5

sau pứ        0              0,1            0,2                   0,2

Chất dư là NaOH: \(m_{NaOH\left(dư\right)}=0,1.40=4\left(g\right)\)

b) \(m_{dd.sau.pứ}=80+100-0,2.98=160,4\left(g\right)\)

=> \(C\%_{NaOH}=\dfrac{0,1.40}{160,4}.100\%=2,5\%\\ C\%_{Na_2SO_4}=\dfrac{0,2.142}{160,4}.100\%=17,71\%\)

a, Gọi: \(\left\{{}\begin{matrix}n_{Al}=x\left(mol\right)\\n_{Mg}=y\left(mol\right)\end{matrix}\right.\)

⇒ 27x + 24y = 13,2 (1)

Ta có: \(n_{H_2}=0,7\left(mol\right)\)

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

\(Mg+2HCl\rightarrow MgCl_2+H_2\)

Theo PT: \(n_{H_2}=\dfrac{3}{2}n_{Al}+n_{Mg}=\dfrac{3}{2}x+y=0,7\left(2\right)\)

Từ (1) và (2) \(\Rightarrow\left\{{}\begin{matrix}x=0,4\left(mol\right)\\y=0,1\left(mol\right)\end{matrix}\right.\)

\(\Rightarrow\left\{{}\begin{matrix}m_{Al}=0,4.27=10,8\left(g\right)\\m_{Mg}=0,1.24=2,4\left(g\right)\end{matrix}\right.\)

b, Theo PT: \(n_{HCl}=2n_{H_2}=1,4\left(mol\right)\)

\(\Rightarrow V_{HCl}=\dfrac{1,4}{2}=0,7\left(l\right)\)

c, Theo PT: \(\left\{{}\begin{matrix}n_{AlCl_3}=n_{Al}=0,4\left(mol\right)\\n_{MgCl_2}=n_{Mg}=0,1\left(mol\right)\end{matrix}\right.\)

\(\Rightarrow\left\{{}\begin{matrix}C_{M_{AlCl_3}}=\dfrac{0,4}{0,7}=\dfrac{4}{7}\left(M\right)\\C_{M_{MgCl_2}}=\dfrac{0,1}{0,7}=\dfrac{1}{7}\left(M\right)\end{matrix}\right.\)

Bạn tham khảo nhé!

Tham khảo ít thooi bạn ơi mik báo giáo viên đấy

a, Ta có: \(n_{H_2}=\dfrac{10,08}{22,4}=0,45\left(mol\right)\)

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

___0,3_____0,9____0,3____0,45 (mol)

\(m_{Al}=0,3.27=8,1\left(g\right)\)

\(\Rightarrow\%m_{Cu}=\dfrac{20-8,1}{20}.100\%=59,5\%\)

b, Có: \(m_{ddHCl}=\dfrac{0,9.36,5}{10\%}=328,5\left(g\right)\)

m _{dd sau pư} = 8,1 + 328,5 - 0,45.2 = 335,7 (g)

\(\Rightarrow C\%_{AlCl_3}=\dfrac{0,3.133,5}{335,7}.100\%\approx11,93\%\)

Bạn tham khảo nhé!

\(n_{Fe}=\dfrac{2,8}{56}=0,05\left(mol\right)\)

PTHH: \(Fe+H_2SO_4\rightarrow FeSO_4+H_2\)

          0,05--->0,05----->0,05--->0,05

\(m_{H_2SO_4}=0,05.98=4,9\left(g\right)\Rightarrow C\%_{H_2SO_4}=\dfrac{4,9}{250}.100\%=1,96\%\)

\(V_{dd.sau.pư}=\dfrac{0,05}{2}=0,025\left(l\right)\)

\(V_{H_2}=0,05.22,4=1,12\left(l\right)\)

\(n_{Fe}=\dfrac{2,8}{56}=0,05\left(mol\right)\)

PTHH: \(Fe+H_2SO_4\rightarrow FeSO_4+H_2\)

         0,05--->0,05------->0,05----->0,05

=> \(\left\{{}\begin{matrix}C\%_{H_2SO_4}=\dfrac{0,05.98}{250}.100\%=1,96\%\\V_{dd.FeSO_4}=\dfrac{0,05}{2}=0,025\left(l\right)=25\left(ml\right)\\V_{H_2}=0,05.22,4=1,12\left(l\right)\end{matrix}\right.\)