Gọi số mol Al, Mg là a, b (mol)

=> 27a + 24b = 7,5 (1)

\(n_{O_2}=\dfrac{3,92}{22,4}=0,175\left(mol\right)\)

PTHH: 4Al + 3O₂ --t^o--> 2Al₂O₃

              a-->0,75a

            2Mg + O₂ --t^o--> 2MgO

               b--->0,5b

=> 0,75a + 0,5b = 0,175 (2)

(1)(2) => a = 0,1 (mol); b = 0,2 (mol)

\(\left\{{}\begin{matrix}\%m_{Al}=\dfrac{0,1.27}{7,5}.100\%=36\%\\\%m_{Mg}=\dfrac{0,2.24}{7,5}.100\%=64\%\end{matrix}\right.\)

a, Ta có: \(n_{Fe}=\dfrac{33,6}{56}=0,6\left(mol\right)\)

\(n_{O_2}=\dfrac{12,395}{24,79}=0,5\left(mol\right)\)

PT: \(3Fe+2O_2\underrightarrow{t^o}Fe_3O_4\)

Xét tỉ lệ: \(\dfrac{0,6}{3}< \dfrac{0,5}{2}\), ta được O₂ dư.

Theo PT: \(n_{Fe_3O_3}=\dfrac{1}{3}n_{Fe}=0,2\left(mol\right)\)

\(\Rightarrow m_{Fe_3O_4}=0,2.232=46,4\left(g\right)\)

b, \(n_{H_2}=\dfrac{5,4198.10^{23}}{6,022.10^{23}}=0,9\left(mol\right)\)

PT: \(Fe_3O_4+4H_2\underrightarrow{t^o}3Fe+4H_2O\)

Xét tỉ lệ: \(\dfrac{0,2}{1}< \dfrac{0,9}{4}\), ta được H₂ dư.

Theo PT: \(n_{Fe}=3n_{Fe_3O_4}=0,6\left(mol\right)\)

\(\Rightarrow m_{Fe}=0,6.56=33,6\left(g\right)\)

Câu 8:

Ta có: \(n_{H_2}=\dfrac{2,24}{22,4}=0,1\left(mol\right)\)

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

PT: \(2H_2+O_2\underrightarrow{t^o}2H_2O\)

Xét tỉ lệ: \(\dfrac{0,1}{2}< \dfrac{0,2}{1}\), ta được O₂ dư.

Theo PT: \(\left\{{}\begin{matrix}n_{O_2\left(pư\right)}=\dfrac{1}{2}n_{H_2}=0,05\left(mol\right)\\n_{H_2O}=n_{H_2}=0,1\left(mol\right)\end{matrix}\right.\)

\(\Rightarrow n_{O_2\left(dư\right)}=0,15\left(mol\right)\)

\(\Rightarrow V_{O_2\left(dư\right)}=0,15.22,4=3,36\left(l\right)\)

\(m_{H_2O}=0,1.18=1,8\left(g\right)\)

Bạn tham khảo nhé!

Câu 9:

a, PT: \(2R+O_2\underrightarrow{t^o}2RO\)

Theo ĐLBT KL, có: m_R + m_O2 = m_RO

⇒ m_O2 = 4,8 (g)

\(\Rightarrow n_{O_2}=\dfrac{4,8}{32}=0,15\left(mol\right)\)

\(\Rightarrow V_{O_2}=0,15.22,4=3,36\left(l\right)\)

b, Theo PT: \(n_R=2n_{O_2}=0,3\left(mol\right)\)

\(\Rightarrow M_R=\dfrac{19,2}{0,3}=64\left(g/mol\right)\)

Vậy: M là đồng (Cu).

Câu 10:

Ta có: m_BaCl2 = 200.15% = 30 (g)

a, m _dd  = 200 + 100 = 300 (g)

\(\Rightarrow C\%_{BaCl_2}=\dfrac{30}{300}.100\%=10\%\)

⇒ Nồng độ dung dịch giảm 5%

b, Ta có: \(C\%_{BaCl_2}=\dfrac{30}{150}.100\%=20\%\)

⇒ Nồng độ dung dịch tăng 5%.

Bạn tham khảo nhé!

Bài 1:

\(a,2Cu+O_2\underrightarrow{t^o}2CuO\)

b, \(n_{O_2}=\dfrac{1,12}{32}=0,035mol\)

\(n_{Cu}=\dfrac{6,4}{64}=0,1mol\)

\(\dfrac{0,1}{2}>\dfrac{0,035}{1}\) => Cu dư, O₂ đủ

\(n_{Cu}\left(dư\right)=0,1-0,07=0,039\left(mol\right)\)

c, \(m_{CuO}=0,07.80=5,6g\)

Bài 2:

\(n_{Al}=\dfrac{13,5}{27}=0,5mol\)

\(n_{O_2}=\dfrac{6,67}{32}=0,21\left(mol\right)\)

\(4Al+3O_2\underrightarrow{t^o}2Al_2O_3\)

\(\dfrac{0,5}{4}>\dfrac{0,21}{3}\) => Al dư, O₂ đủ

\(n_{Al_2O_3}=\dfrac{2}{3}.0,21=0,14\left(mol\right)\)

\(m_{Al_2O_3}=0,14.102=14,28g\)

\(n_{O_2}=\dfrac{6,72}{22,4}=0,3\left(mol\right)\)

Theo ĐLBTKL: m_M + m_O2 = m_MxOy

=> m_M = 20,4 - 0,3.32 = 10,8(g)

\(n_M=\dfrac{10,8}{M_M}\left(mol\right)\)

PTHH: 2xM + yO₂ --t^o--> 2M_xO_y

_____\(\dfrac{10,8}{M_M}\) ->\(\dfrac{10,8y}{2x.M_M}\)

=>\(\dfrac{10,8y}{2x.M_M}=0,3\)

=> \(M_M=9.\dfrac{2y}{x}\)

Xét \(\dfrac{2y}{x}=1=>L\)

Xét \(\dfrac{2y}{x}=2=>L\)

Xét \(\dfrac{2y}{x}=3=>M_M=27\left(Al\right)\)

a. \(n_{O_2}=\dfrac{3.36}{22,4}=0,15\left(mol\right)\)

PTHH : 2Zn + O₂ -------t^o------> 2ZnO

             0,3      0,15                        0,15

\(m_{Zn}=65.0,3=19,5\left(g\right)\)

b. \(m_{ZnO}=0,15.81=12,15\left(g\right)\)

c. PTHH : 2KMnO₄ ---t^o---> K₂MnO₄ + MnO₂ + O₂

                  0,3                                                    0,15

\(m_{KMnO_4}=158.0,3=47,4\left(g\right)\)

\(n_{O_2}=\dfrac{33,6}{22,4}=1,5\left(mol\right)\)

PTHH: 2Zn + O₂ --to--> 2ZnO

             3      1,5               3

\(\rightarrow\left\{{}\begin{matrix}m_{Zn}=3.65=195\left(g\right)\\m_{ZnO}=81.3=243\left(g\right)\end{matrix}\right.\)

2KMnO₄ --to--> K₂MnO₄ + MnO₂ + O₂

6                                                        3

=> m_KMnO4 = 6.158 = 948 (g)

nO₂ = \(\dfrac{6,72}{22,4}=0,3\left(mol\right)\)

pt: \(2KMnO_4\rightarrow K_2MnO_4+MnO_2+O_2\)

Theo pt: \(n_{KMnO_4}=2n_{O_2}=0,6mol\)

=> nKMnO₄ thực tế = 0,6:\(\dfrac{90}{100}=\dfrac{2}{3}\left(mol\right)\)

mKMnO₄ = \(\dfrac{2}{3}.158=\dfrac{316}{3}g\)

\(a,PTHH:2X+6HCl\to 2XCl_3+3H_2\\ b,n_{H_2}=\dfrac{6,72}{22,4}=0,3(mol)\\ \Rightarrow n_{X}=\dfrac{2}{3}n_{H_2}=0,2(mol)\\ \Rightarrow M_{X}=\dfrac{5,4}{0,2}=27(g/mol)\)

Vậy X là nhôm (Al)

\(c,n_{AlCl_3}=n_{Al}=0,2(mol)\\ \Rightarrow m_{AlCl_3}=0,2.133,5=26,7(g)\)

TN1: Gọi (n_Cu, n_Al, n_Fe) = (a,b,c)

=> 64a + 27b + 56c = 14,3 (1)

\(n_{H_2}=\dfrac{6,72}{22,4}=0,3\left(mol\right)\)

PTHH: 2Al + 6HCl --> 2AlCl₃ + 3H₂

             b----------------------->1,5b

            Fe + 2HCl --> FeCl₂ + H₂

              c----------------------->c

=> 1,5b + c = 0,3 (2)

TN2: Gọi (n_Cu, n_Al, n_Fe) = (ak,bk,ck)

=> ak + bk + ck = 0,6 (3)

\(n_{O_2}=\dfrac{44,8}{22,4}.20\%=0,4\left(mol\right)\)

PTHH: 2Cu + O₂ --t^o--> 2CuO

            ak--->0,5ak

            4Al + 3O₂ --t^o--> 2Al₂O₃

           bk--->0,75bk

            3Fe + 2O₂ --t^o--> Fe₃O₄

           ck-->\(\dfrac{2}{3}ck\)

=> 0,5ak + 0,75bk + \(\dfrac{2}{3}ck\) = 0,4 (4)

(1)(2)(3) => \(\left\{{}\begin{matrix}a=0,05\left(mol\right)\\b=0,1\left(mol\right)\\c=0,15\left(mol\right)\\k=2\end{matrix}\right.\)

=> \(\left\{{}\begin{matrix}\%m_{Cu}=\dfrac{0,05.64}{14,3}.100\%=22,38\%\\\%m_{Al}=\dfrac{0,1.27}{14,3}.100\%=18,88\%\\\%m_{Fe}=\dfrac{0,15.56}{14,3}.100\%=58,74\%\end{matrix}\right.\)