oxygen  nx a

Gọi công thức phân tử của X là: \({\rm{ }}{{\rm{C}}_{\rm{x}}}{{\rm{H}}_{\rm{y}}}{{\rm{O}}_z}\)

\({\rm{\% }}{{\rm{m}}_{\rm{O}}}{\rm{  =  100\%   -  (52,17\%   +  13,04\%  )  =  34,79\% }}\)

Từ phổ khối lượng, ta có M = 46 (amu)

\(\begin{array}{l}{\rm{x  =  }}\frac{{{\rm{\% }}{{\rm{m}}_{\rm{C}}}}}{{{\rm{12}}}}{\rm{ \times }}\frac{{\rm{M}}}{{{\rm{100}}}}{\rm{  =  }}\frac{{{\rm{52,17}}}}{{{\rm{12}}}}{\rm{ \times }}\frac{{{\rm{46}}}}{{{\rm{100}}}} \approx 2\\{\rm{y  =  }}\frac{{{\rm{\% }}{{\rm{m}}_{\rm{H}}}}}{{\rm{1}}}{\rm{ \times }}\frac{{\rm{M}}}{{{\rm{100}}}}{\rm{ =  }}\frac{{{\rm{13,04}}}}{{\rm{1}}}{\rm{ \times }}\frac{{46}}{{{\rm{100}}}} \approx {\rm{6}}\\{\rm{z  =  }}\frac{{{\rm{\% }}{{\rm{m}}_{\rm{O}}}}}{{{\rm{16}}}}{\rm{ \times }}\frac{{\rm{M}}}{{{\rm{100}}}}{\rm{ =  }}\frac{{34,79}}{{{\rm{16}}}}{\rm{ \times }}\frac{{46}}{{{\rm{100}}}} \approx {\rm{1}}\end{array}\)

Công thức phân tử của X là\({\rm{ }}{{\rm{C}}_{\rm{2}}}{{\rm{H}}_{\rm{6}}}{\rm{O }}\)

PTK_naphthalene = 128

PTK_phenol = 94

\(2CO+O_2\rightarrow\left(t^o\right)2CO_2\\ n_{CO}=\dfrac{2,479}{24,79}=0,1\left(mol\right)\\ n_{O_2}=\dfrac{0,1}{2}=0,05\left(mol\right)\\ V_{O_2\left(đkc\right)}=0,05.24,79=1,2395\left(l\right)\)

mình cùng ko  bt thế này đúng  ko

2CO+O2→(to)2CO2nCO=2,47924,79=0,1(mol)nO2=0,12=0,05(mol)VO2(đkc)=0,05.24,79=1,2395(l)

Bài 3:

\(n_{Al}=\dfrac{5,4}{27}=0,2\left(mol\right);n_{H_2SO_4}=0,45.1=0,45\left(mol\right)\\ PTHH:2Al+3H_2SO_4\rightarrow Al_2\left(SO_4\right)_3+3H_2\\ Vì:\dfrac{0,2}{2}< \dfrac{0,45}{3}\Rightarrow H_2SO_4dư\\ n_{Al_2\left(SO_4\right)_3}=\dfrac{n_{Al}}{2}=\dfrac{0,2}{2}=0,1\left(mol\right)\\ m_{muối}=m_{Al_2\left(SO_4\right)_3}=342.0,1=34,2\left(g\right)\)

Bài 2:

\(n_{Zn}=\dfrac{3,25}{65}=0,05\left(mol\right);n_{HCl}=0,12.1=0,12\left(mol\right)\\ Zn+2HCl\rightarrow ZnCl_2+H_2\\ Vì:\dfrac{0,05}{1}< \dfrac{0,12}{2}\Rightarrow HCldư\\ n_{H_2}=n_{Zn}=0,05\left(mol\right)\\ V_{H_2\left(đkc\right)}=0,05.24,79=1,2395\left(l\right)\)

\(n_{Fe_2O_3}=\dfrac{12}{160}=0,075\left(mol\right)\\a, Fe_2O_3+3H_2\rightarrow\left(t^o\right)2Fe+3H_2O\\ b,n_{H_2}=3.0,075=0,225\left(mol\right)\\ V_{H_2\left(đkc\right)}=24,79.0,225=5,57775\left(l\right)\\ c,n_{Fe}=2.0,075=0,15\left(mol\right)\\ m_{Fe}=0,15.56=8,4\left(g\right)\)

Câu 381:

\(n_{Al}=\dfrac{12,42}{27}=0,46\left(mol\right)\)

\(\left\{{}\begin{matrix}n_{N_2O}+n_{N_2}=\dfrac{2,688}{22,4}=0,12\left(mol\right)\\44n_{N_2O}+28n_{N_2}=18.2.0,12\end{matrix}\right.\)

\(\Rightarrow\left\{{}\begin{matrix}n_{N_2O}=0,06\left(mol\right)\\n_{N_2}=0,06\left(mol\right)\end{matrix}\right.\)

BT e, có: \(3n_{Al}=8n_{N_2O}+10n_{N_2}+8n_{NH_4NO_3}\)

\(\Rightarrow n_{NH_4NO_3}=0,0375\left(mol\right)\)

1. BTNT Al: n_Al(NO3)3 = n_Al = 0,46 (mol)

⇒ m = 0,46.213 + 0,0375.80 = 100,98 (g)

→ Đáp án: D

2. n_HNO3 = 10n_N2O + 12n_N2 + 10n_NH4NO3 = 1,695 (mol)

→ Đáp án: A

Câu 378:

\(n_{Al}=\dfrac{10,8}{27}=0,4\left(mol\right)\)

\(n_{N_2}=\dfrac{1,792}{22,4}=0,08\left(mol\right)\)

BT e, có: \(3n_{Al}=10n_{N_2}+8n_{NH_4NO_3}\)

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

1. BTNT Al: n_Al(NO3)3 = n_Al = 0,4 (mol)

⇒ m = m_Al(NO3)3 + m_NH4NO3 = 0,4.213 + 0,05.80 = 89,2 (g)

→ Đáp án: D

2. n_HNO3 = 12n_N2 + 10n_NH4NO3 = 1,46 (mol)

\(\Rightarrow V_{HNO_3}=\dfrac{1,46}{2}=0,73\left(l\right)=730\left(ml\right)\)

Câu 379:

\(n_{Al}=\dfrac{12,42}{27}=0,46\left(mol\right)\)

Ta có: \(\left\{{}\begin{matrix}n_{N_2O}+n_{N_2}=\dfrac{1,344}{22,4}=0,06\left(mol\right)\\44n_{N_2O}+28n_{N_2}=18.2.0,06\end{matrix}\right.\)

\(\Rightarrow\left\{{}\begin{matrix}n_{N_2O}=0,03\left(mol\right)\\n_{N_2}=0,03\left(mol\right)\end{matrix}\right.\)

BT e, có: \(3n_{Al}=8n_{N_2O}+10n_{N_2}+8n_{NH_4NO_3}\)

\(\Rightarrow n_{NH_4NO_3}=0,105\left(mol\right)\)

1. BTNT Al: n_Al(NO3)3 = n_Al = 0,46 (mol)

⇒ m = m_Al(NO3)3 + m_NH4NO3 = 0,46.213 + 0,105.80 = 106,38 (g)

→ Đáp án: C

2. n_HNO3 = 10n_N2O + 12n_N2 + 10n_NH4NO3 = 1,71 (mol)

→ Đáp án: A

Câu 356:

Ta có: 56n_Fe + 64n_Cu + 108n_Ag = 54,28 (1)

BTNT Fe: n_Fe2(SO4)3 = 1/2n_Fe

BTNT Cu: n_CuSO4 = n_Cu

⇒ 200n_Fe - 160n_Cu = 27,2 (2)

\(m_{H_2SO_4}=179.98\%=175,42\left(g\right)\Rightarrow n_{H_2SO_4}=\dfrac{175,42}{98}=1,79\left(mol\right)\)

Có: n_H2SO4 = 2n_SO2 ⇒ n_SO2 = 0,895 (mol)

BT e, có: 3n_Fe + 2n_Cu + n_Ag = 2n_SO2 = 1,79 (3)

Từ (1), (2) và (3) \(\Rightarrow\left\{{}\begin{matrix}n_{Fe}=0,36\left(mol\right)\\n_{Cu}=0,28\left(mol\right)\\n_{Ag}=0,15\left(mol\right)\end{matrix}\right.\)

\(\Rightarrow\%m_{Ag}=\dfrac{0,15.108}{54,28}.100\%\approx29,85\%\)

→ Đáp án: C

Câu 357:

Ta có: \(65n_{Zn}+56n_{Fe}+64n_{Cu}=64,34\left(1\right)\)

\(n_{H_2}=\dfrac{15,008}{22,4}=0,67\left(mol\right)\)

BT e, có: \(2n_{Zn}+2n_{Fe}=2n_{H_2}=1,34\left(2\right)\)

\(m_{H_2SO_4}=231.98\%=226,38\left(g\right)\Rightarrow n_{H_2SO_4}=\dfrac{226,38}{98}=2,31\left(mol\right)\)

Mà: n_H2SO4 = 2n_SO2 ⇒ n_SO2 = 1,155 (mol)

BT e, có: 2n_Zn + 3n_Fe + 2n_Cu = 2n_SO2 = 2,31 (3)

Từ (1), (2) và (3) \(\Rightarrow\left\{{}\begin{matrix}n_{Zn}=0,42\left(mol\right)\\n_{Fe}=0,25\left(mol\right)\\n_{Cu}=0,36\left(mol\right)\end{matrix}\right.\)

\(\Rightarrow\%m_{Zn}=\dfrac{0,42.65}{64,34}.100\%\approx42,43\%\)

→ Đáp án: C

Ta có: 27n_Al + 56n_Fe + 64n_Cu = 23,82 (1)

\(n_{H_2}=\dfrac{9,408}{22,4}=0,42\left(mol\right)\)

BT e, có: 3n_Al + 2n_Fe = 2n_H2 = 0,42.2 (2)

Có: n_HNO3 = 0,8.2 = 1,6 (mol)

\(\Rightarrow\left\{{}\begin{matrix}n_{N_2}+n_{NO}=\dfrac{3,584}{22,4}=0,16\\12n_{N_2}+4n_{NO}=n_{HNO_3}=1,6\end{matrix}\right.\)

\(\Rightarrow\left\{{}\begin{matrix}n_{N_2}=0,12\left(mol\right)\\n_{NO}=0,04\left(mol\right)\end{matrix}\right.\)

BT e, có: 3n_Al + 3n_Fe + 2n_Cu = 10n_N2 + 3n_NO = 1,32 (3)

Từ (1), (2) và (3) \(\Rightarrow\left\{{}\begin{matrix}n_{Al}=0,18\left(mol\right)\\n_{Fe}=0,15\left(mol\right)\\n_{Cu}=0,165\left(mol\right)\end{matrix}\right.\)

\(\Rightarrow\%m_{Al}=\dfrac{0,18.27}{23,82}.100\%\approx20,4\%\)

→ Đáp án: B