Câu 2:

a, \(2NaOH+H_2SO_4\rightarrow Na_2SO_4+2H_2O\)

c, \(n_{NaOH}=\dfrac{40}{40}=1\left(mol\right)\)

Theo PT: \(n_{H_2SO_4}=\dfrac{1}{2}n_{NaOH}=0,5\left(mol\right)\)

\(\Rightarrow V_{H_2SO_4}=\dfrac{0,5}{2}=0,25\left(l\right)\)

d, \(n_{Na_2SO_4}=\dfrac{1}{2}n_{NaOH}=0,5\left(mol\right)\)

\(\Rightarrow C_{M_{Na_2SO_4}}=\dfrac{0,5}{0,25}=2\left(M\right)\)

Câu 1:

a, \(Zn+2HCl\rightarrow ZnCl_2+H_2\)

b, \(n_{Zn}=\dfrac{16,25}{65}=0,25\left(mol\right)\)

\(n_{H_2}=n_{Zn}=0,25\left(mol\right)\)

\(\Rightarrow V_{H_2}=0,25.24,79=6,1975\left(l\right)\)

c, \(n_{HCl}=2n_{Zn}=0,5\left(mol\right)\)

\(\Rightarrow m_{HCl}=0,5.36,5=18,25\left(g\right)\)

\(\Rightarrow m_{ddHCl}=\dfrac{18,25}{10\%}=182,5\left(g\right)\)

d, \(n_{ZnCl_2}=n_{Zn}=0,25\left(mol\right)\)

\(\Rightarrow C\%_{ZnCl_2}=\dfrac{0,25.136}{16,25+182,5-0,25.2}.100\%\approx17,15\%\)

Ta có: m_{KL (trong 1 phần)} = 6,48:2 = 3,24 

- Phần 1:

m _muối = m_KL + m_Cl^- \(\Rightarrow n_{Cl^-}=\dfrac{16,02-3,24}{35,5}=0,36\left(mol\right)\)

BTNT Cl: n_HCl = n_Cl^- = 0,36 (mol)

BTNT H: n_H2 = 1/2n_HCl = 0,18 (mol)

\(\Rightarrow V_{H_2}=0,18.22,4=4,032\left(l\right)\)

- Phần 2: 

BT e, có: 2n_Zn + 2n_Mg + 3n_Al = 2n_H2 = 4n_O2

⇒ n_O2 = 0,09 (mol)

⇒ m _oxide = m_KL + m_O2 = 3,24 + 0,09.32 = 6,12 (g)

Ta có: 56n_Fe + 88n_FeS = 20 (1)

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

BT e: 2n_H2 = 2n_Fe ⇒ n_H2 = n_Fe

BTNT S: n_H2S = n_FeS

⇒ n_Fe + n_FeS = 0,3 (2)

Từ (1) và (2) \(\Rightarrow\left\{{}\begin{matrix}n_{Fe}=0,2\left(mol\right)\\n_{FeS}=0,1\left(mol\right)\end{matrix}\right.\)

Coi A gồm: Fe: 0,3 (mol) và S: 0,1 (mol)

BT e, có: 2n_Fe + 4n_S = 2n_SO2

⇒ n_SO2 = 0,5 (mol)

BTNT Fe: n_FeSO4 = n_Fe = 0,3 (mol)

BTNT S: n_H2SO4 = 0,3 + 0,5 = 0,8 (mol)

a, \(4P+5O_2\underrightarrow{t^o}2P_2O_5\)

b, \(MgO+2HCl\rightarrow MgCl_2+H_2O\)

c, \(Ba\left(OH\right)_2+H_2SO_4\rightarrow BaSO_4+2H_2O\)

d, \(CaCO_3+2HCl\rightarrow CaCl_2+CO_2+H_2O\)

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

f, \(CuO+2HCl\rightarrow CuCl_2+H_2O\)

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

h, \(Fe+CuSO_4\rightarrow FeSO_4+Cu\)

Câu 385:

Ta có: \(n_{Mg}=\dfrac{4,8}{24}=0,2\left(mol\right)=n_{Mg\left(NO_3\right)_2}\)

\(n_{Zn}=\dfrac{13}{65}=0,2\left(mol\right)=n_{Zn\left(NO_3\right)_2}\)

\(n_{N_2O}=\dfrac{1,12}{22,4}=0,05\left(mol\right)\)

1. BT e, có: \(2n_{Mg}+2n_{Zn}=8n_{N_2O}+8n_{NH_4NO_3}\)

\(\Rightarrow n_{NH_4NO_3}=0,05\left(mol\right)\)

\(\Rightarrow m_2=m_{Mg\left(NO_3\right)_2}+m_{Zn\left(NO_3\right)_2}+m_{NH_4NO_3}=71,4\left(g\right)\)

→ Đáp án: C

2. n_HNO3 = 10n_N2O + 10n_NH4NO3 = 1 (mol)

\(\Rightarrow m_1=\dfrac{1.63}{12,6\%}=500\left(g\right)\)

→ Đáp án: C

a, \(Zn+2HCl\rightarrow ZnCl_2+H_2\)

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

b, Ta có: \(n_{H_2}=\dfrac{5,6}{22,4}=0,25\left(mol\right)\)

Theo PT: \(n_{HCl}=2n_{H_2}=0,5\left(mol\right)\)

\(\Rightarrow C_{M_{HCl}}=\dfrac{0,5}{0,1}=5\left(M\right)\)

c, Ta có: 65n_Zn + 24n_Mg = 10,1 (1)

Theo PT: \(n_{H_2}=n_{Zn}+n_{Mg}=0,25\left(2\right)\)

Từ (1) và (2) \(\Rightarrow\left\{{}\begin{matrix}n_{Zn}=0,1\left(mol\right)\\n_{Mg}=0,15\left(mol\right)\end{matrix}\right.\)

\(\Rightarrow\left\{{}\begin{matrix}m_{Zn}=0,1.65=6,5\left(g\right)\\m_{Mg}=0,15.24=3,6\left(g\right)\end{matrix}\right.\)

Câu 384:

Ta có: \(n_{Al}=\dfrac{5,4}{27}=0,2\left(mol\right)\)

\(n_{Zn}=\dfrac{19,5}{65}=0,3\left(mol\right)\)

\(n_{N_2O}=\dfrac{2,24}{22,4}=0,1\left(mol\right)\)

1. BT e, có: 3n_Al + 2n_Zn = 8n_N2O + 8n_NH4NO3

⇒ n_NH4NO3 = 0,05 (mol)

BTNT Al: n_Al(NO3)3 = n_Al = 0,2 (mol)

BTNT Zn: n_Zn(NO3)2 = n_Zn = 0,3 (mol)

⇒ m₂ = m_Al(NO3)3 + m_Zn(NO3)2 + m_NH4NO3 = 103,3 (g)

→ Đáp án: C

2. Có: n_HNO3 = 10n_N2O + 10n_NH4NO3 = 1,5 (mol)

\(\Rightarrow m_{ddHNO_3}=\dfrac{1,5.63}{12,6\%}=750\left(g\right)\)

→ Đáp án: A

Câu 388 mình đã làm ở trên rồi nhé.

Câu 387:

Ta có: \(n_{Mg}=\dfrac{3,6}{24}=0,15\left(mol\right)\)

\(n_{Al}=\dfrac{8,1}{27}=0,3\left(mol\right)\)

Có: \(\left\{{}\begin{matrix}n_{N_2}+n_{N_2O}=\dfrac{1,792}{22,4}=0,08\\28n_{N_2}+44n_{N_2O}=0,08.18.2\end{matrix}\right.\) \(\Rightarrow\left\{{}\begin{matrix}n_{N_2}=0,04\left(mol\right)\\n_{N_2O}=0,04\left(mol\right)\end{matrix}\right.\)

BT e, có: 2n_Mg + 3n_Al = 10n_N2 + 8n_N2O + 8n_NH4NO3

⇒ n_NH4NO3 = 0,06 (mol)

1. BTNT Mg: n_Mg(NO3)2 = n_Mg = 0,15 (mol)

BTNT Al: n_Al(NO3)3 = n_Al = 0,3 (mol)

⇒ m = m_Mg(NO3)2 + m_Al(NO3)3 + m_NH4NO3 = 90,9 (g)

→ Đáp án: D

2. n_HNO3 = 12n_N2 + 10n_N2O + 10n_NH4NO3 = 1,48 (mol)

\(\Rightarrow m_{ddHNO_3}=\dfrac{1,48.63}{15,75\%}=592\left(g\right)\)

→ Đáp án: D