\(n_{FeO}=\dfrac{2,16}{72}=0,03\left(mol\right)\\ n_{HCl}=0,2.0,4=0,08\left(mol\right)\\ a,PTHH:FeO+2HCl\rightarrow FeCl_2+H_2O\\ Vì:\dfrac{0,03}{1}< \dfrac{0,08}{2}\Rightarrow HCldư\\ n_{FeCl_2}=n_{FeO}=0,03\left(mol\right)\\ m_{FeCl_2}=127.0,03=3,81\left(g\right)\\ n_{HCl\left(Dư\right)}=0,08-2.0,03=0,02\left(mol\right)\\ V_{ddsau}=V_{ddHCl}=0,4\left(l\right)\\ C_{MddHCl\left(dư\right)}=\dfrac{0,02}{0,4}=0,05\left(M\right);C_{MddFeCl_2}=\dfrac{0,03}{0,4}=0,075\left(M\right)\)

\(n_{CuO}=\dfrac{4}{80}=0,05\left(mol\right)\\a, CuO+H_2SO_4\rightarrow CuSO_4+H_2O\\ n_{CuSO_4}=n_{H_2SO_4}=n_{CuO}=0,05\left(MOL\right)\\ b,m_{CuSO_4}=0,05.160=8\left(g\right)\\ c,V_{ddH_2SO_4}=\dfrac{0,05}{0,5}=0,1\left(l\right)\\ d,V_{ddCuSO_4}=V_{ddH_2SO_4}=0,1\left(l\right)\\ C_{MddCuSO_4}=\dfrac{0,05}{0,1}=0,5\left(M\right)\)

\(a)4Al+3O_2\xrightarrow[]{t^0}2Al_2O_3\\ b)n_{Al_2O_3}=\dfrac{20,4}{102}=0,2mol\\ n_{Al}=\dfrac{0,2.4}{2}=0,4mol\\ m_{Al}=0,4.27=10,8g\)

c) bạn xem lại đề

\(a)7437ml=7,437l\\ n_{N_2}=\dfrac{7,437}{24,79}=0,3mol\\ m_{N_2}=0,3.28=8,4g\\ b)n_{Cl_2}=\dfrac{2,479}{24,79}=0,1mol\\ m_{Cl_2}=0,1.71=7,1g\)

b, 14,9 g

Oxit: MO.

Gọi: n_CuO = a (mol) → n_MO = 2a (mol)

⇒ 80a + (M_M + 16).2a = 3,6 (1)

Có: n_HNO3 = 0,06.2,5 = 0,15 (mol)

BTNT Cu: n_Cu(NO3)2 = n_Cu = a (mol)

BTNT M: n_M(NO3)2 = n_M = 2a (mol)

BT e, có: 2n_Cu = 3n_NO \(\Rightarrow n_{NO}=\dfrac{2}{3}n_{Cu}=\dfrac{2}{3}a\left(mol\right)\)

BTNT N, có: \(2n_{Cu\left(NO_3\right)_2}+2n_{M\left(NO_3\right)_2}+n_{NO}=n_{HNO_3}\)

\(\Rightarrow2a+2.2a+\dfrac{2}{3}a=0,15\left(2\right)\)

Từ (1) và (2) \(\Rightarrow\left\{{}\begin{matrix}a=0,0225\left(mol\right)\\M_M=24\left(g/mol\right)\end{matrix}\right.\)

→ M là Mg.

Đáp án: B

Ta có: \(n_{H_2}=\dfrac{2,479}{24,79}=0,1\left(mol\right)\)

Theo ĐLBT KL: m_Zn + m_H2SO4 = m_ZnSO4 + m_H2

⇒ m_H2SO4 = 9 + 0,1.2 - 6,5 = 2,7 (g)

a, \(N_2+3H_2⇌2NH_3\)

b, Ta có: \(K_c=\dfrac{\left[NH_3\right]^2}{\left[N_2\right]\left[H_2\right]^3}=\dfrac{0,62^2}{0,45.0,14^3}\approx311,31\)

c, - Tăng nhiệt độ → giảm hiệu suất.

- Tăng áp suất → tăng hiệu suất.

- Thêm bột sắt (xúc tác) → không làm thay đổi hiệu suất.

Ta có: \(n_{KOH}=0,05.0,1=0,005\left(mol\right)\)

\(n_{HNO_3}=0,052.0,1=0,0052\left(mol\right)\)

PT: \(KOH+HNO_3\rightarrow KNO_3+H_2O\)

Xét tỉ lệ: \(\dfrac{0,005}{1}< \dfrac{0,0052}{1}\), ta được HNO₃ dư.

Theo PT: \(n_{HNO_3\left(pư\right)}=n_{KOH}=0,005\left(mol\right)\)

\(\Rightarrow n_{H^+}=n_{HNO_3\left(dư\right)}=0,0002\left(mol\right)\)

\(\Rightarrow\left[H^+\right]=\dfrac{0,0002}{0,05+0,052}=\dfrac{1}{510}\left(M\right)\)

⇒ pH = -log[H⁺] ≃ 2,71