D

Bài 8:

\(n_{Al}=\dfrac{2,7}{27}=0,1\left(mol\right)\)

\(2Al+3H_2SO_4\rightarrow Al_2\left(SO_4\right)_3+3H_2\)

0,1--->0,15------>0,05---------->0,15

a. \(V_{H_2SO_4}=\dfrac{0,15}{1}=0,15\left(l\right)=150\left(ml\right)\)

b. \(V'=0,15.22,4=3,36\left(l\right)\)

c. \(CM_{Al_2\left(SO_4\right)_3}=\dfrac{0,05}{0,15}=\dfrac{1}{3}M\)

Câu 2

\((1) MnO_2 + 4HCl \to MnCl_2 + Cl_2 + 2H_2O\\ (2) Cl_2 + H_2 \xrightarrow{as} 2HCl\\ (3) 3Cl_2 + 2Fe \xrightarrow{t^o} 2FeCl_3\\ (4) 2FeCl_3 + Fe \to 3FeCl_2\\ (5) 2NaOH + Cl_2 \to NaCl + NaClO + H_2O\)

\((1) 4Al + 3O_2 \xrightarrow{t^o} 2Al_2O_3\\ (2) 2Fe + 3Cl_2 \xrightarrow{t^o} 2FeCl_3\\ (3) C + O_2 \xrightarrow{t^o} CO_2\\ (4) 2KMnO_4 \xrightarrow{t^o} K_2MnO_4 + MnO_2 + O_2\\ (5) 4P + 5O_2 \xrightarrow{t^o} 2P_2O_5\\ (6) 2KClO_3 \xrightarrow{t^o} 2KCl + 3O_2\\ (7) Fe + H_2SO_4 \to FeSO_4 + H_2\\ (8) Cu + 2H_2SO_4 \to CuSO_4 + SO_2 + 2H_2O\\ (9) 2Fe + 6H_2SO_4 \to Fe_2(SO_4)_3 + 3SO_2 + 6H_2O\\ (10) 2Al + 6H_2SO_4 \to Al_2(SO_4)_3 + 3SO_2 + 6H_2O\)

Sửa đề là : 4.6 (g) 

\(n_{H_2}=\dfrac{2.24}{22.4}=0.1\left(mol\right)\)

\(A+H_2O\rightarrow AOH+\dfrac{1}{2}H_2\)

\(0.2...............................0.1\)

\(M_A=\dfrac{4.6}{0.2}=23\left(\dfrac{g}{mol}\right)\)

\(A:Na\)

Đề này C1 em sửa thành 4,6 gam kim loại như bạn dưới, C2 em sửa thành 22,4 lít H₂

a) \(n_{MgCO_3}=\dfrac{33,6}{84}=0,4\left(mol\right)\)

PTHH: MgCO₃ + 2HNO₃ --> Mg(NO₃)₂ + CO₂ + H₂O

_______0,4---->0,8----------->0,4

\(C_{M\left(HNO_3\right)\left(C\right)}=\dfrac{0,8}{0,5}=1,6M\)

\(C_{M\left(Mg\left(NO_3\right)_2\right)\left(D\right)}=\dfrac{0,4}{0,5}=0,8M\)

b) Gọi \(\left\{{}\begin{matrix}C_{M\left(A\right)}=aM\\C_{M\left(B\right)}=bM\end{matrix}\right.\)

300ml dd A chứa: 0,3a mol HNO₃

200ml dd B chứa: 0,2b mol HNO₃

=> 0,3a + 0,2b = 0,8 (1)

200ml dd A chứa: 0,2a mol HNO₃

300ml dd B chứa: 0,3b mol HNO₃

=> \(\dfrac{0,2a+0,3b}{0,5}=1,4\) (2)

(1)(2) => a = 2 ; b = 1

^{BỚT TRẢ LỜI THIẾU SUY NGHĨ !!}

a) Cl₂ + 2NaOH --> NaClO + NaCl + H₂O

Chất oxh: Cl₂, chất khử: Cl₂

b) \(n_{Cl_2}=\dfrac{17,92}{22,4}=0,8\left(mol\right);n_{NaOH}=0,5.4=2\left(mol\right)\)

PTHH: Cl₂ + 2NaOH --> NaClO + NaCl + H₂O

_____0,8---->1,6--------->0,8---->0,8

=> \(\left\{{}\begin{matrix}C_{M\left(NaCl\right)}=\dfrac{0,8}{0,5}=1,6M\\C_{M\left(NaClO\right)}=\dfrac{0,8}{0,5}=1,6M\\C_{M\left(NaOH\right)}=\dfrac{2-1,6}{0,5}=0,8M\end{matrix}\right.\)

https://hoidap247.com/cau-hoi/558074

tham khảo ạ

nH2SO4=0,045mol    nKOH=0,03

PT: X + H2SO4 ==> XSO4 + H2 (1)

      H2SO4 + 2KOH ==> K2SO4 + 2H2O (2)

Từ 1 --> nH2SO4 dư = 0,015

--> nX=nH2SO4 pứ= 0,045-0,015=0,03

MX=1,2:0,03=40 --> X là Ca