Bài 15:

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

PTHH: 2H₂O --đp--> 2H₂ + O₂

               1<--------------0,5

=> \(H=\dfrac{1.18}{22,5}.100\%=80\%\)

b) \(n_{H_2O}=\dfrac{81}{18}=4,5\left(mol\right)\)

PTHH: 2H₂ + O₂ --t^o--> 2H₂O

           4,5<------------4,5

=> \(V_{H_2\left(lý.thuyết\right)}=4,5.24,79=111,555\left(l\right)\)

=> \(V_{H_2\left(tt\right)}=\dfrac{111,555.100}{90}=123,95\left(l\right)\)

c) \(n_{H_2}=\dfrac{30,9875}{24,79}=1,25\left(mol\right)\)

PTHH: 2H₂O --đp--> 2H₂ + O₂

            1,25<-------1,25

=> \(m_{H_2O\left(lý.thuyết\right)}=1,25.18=22,5\left(g\right)\)

=> \(m_{H_2O\left(tt\right)}=\dfrac{22,5.100}{75}=30\left(g\right)\)

Bài 8

a) \(n_{H_2}=\dfrac{37,185}{24,79}=1,5\left(mol\right)\)

PTHH: Fe₂O₃ + 3H₂ --t^o--> 2Fe + 3H₂O

              0,5<----1,5-------------->1,5

=> m_Fe2O3 = 0,5.150 = 80 (g)

b) m_H2O = 1,5.18 = 27 (g)

=> \(V_{H_2O}=\dfrac{27}{1}=27\left(ml\right)\)

\(a.2H_2+O_2-^{t^o}\rightarrow2H_2O\\ b.n_{H_2}=\dfrac{8,96}{22,4}=0,4\left(mol\right)\\ n_{O_2}=\dfrac{1}{2}n_{H_2}=0,2\left(mol\right)\\ \Rightarrow V_{O_2}=0,2.22,4=4,48\left(l\right)\\ TrongkhôngkhíO_2chiếm20\%\\ \Rightarrow V_{kk}=\dfrac{4,48}{20\%}=22,4\left(l\right)\)

Vì hiệu suất là 25%

=> \(m_{NH_3}=17.\dfrac{25\%}{100\%}=4,25\left(g\right)\)

Ta có: \(n_{NH_3}=\dfrac{4,25}{17}=0,25\left(mol\right)\)

PTHH: N₂ + 3H₂ ---> 2NH₃

Theo PT: \(n_{H_2}=\dfrac{3}{2}.n_{NH_3}=\dfrac{3}{2}.0,25=0,375\left(mol\right)\)

=> \(V_{H_2}=0,375.22,4=8,4\left(lít\right)\)

Theo PT: \(n_{N_2}=\dfrac{1}{2}.n_{NH_3}=\dfrac{1}{2}.0,25=0,125\left(mol\right)\)

=> \(V_{N_2}=0,125.22,4=2,8\left(lít\right)\)

\(n_{NH_3}=1\left(mol\right)\)

Bảo toàn nguyên tố:

\(\left\{{}\begin{matrix}n_{N_2}=\dfrac{n_{NH_3}}{2}=0,5\left(mol\right)\\n_{H_2}=\dfrac{3n_{NH_3}}{2}=1,5\left(mol\right)\end{matrix}\right.\)

Do hiệu suất 25% \(\Rightarrow\left\{{}\begin{matrix}n_{N_2}=2\left(mol\right)\\n_{H_2}=6\left(mol\right)\end{matrix}\right.\Rightarrow\left\{{}\begin{matrix}V_{N_2}=44,8\left(l\right)\\n_{H_2}=134,4\left(l\right)\end{matrix}\right.\)

Theo PTHH

\(n_{H2}=n_{H20}=\dfrac{3,6}{18}=0,2(mol)\)

\(V_{H2}=0,2 . 22,4 =4,48 (lít)\)

Theo PTHH :

a, Theo giả thiết ta có: \(n_P=\dfrac{3,1}{31}=0,1\left(mol\right)\)

\(4P+5O_2--t^o->2P_2O_5\)

Ta có: \(n_{O_2}=\dfrac{5}{4}.n_P=0,125\left(mol\right)\Rightarrow V_{O_2\left(đktc\right)}=0,125.22,4=2,8\left(l\right)\)

b, Theo giả thiết ta có: \(n_{CH_4}=\dfrac{1,12}{22,4}=0,05\left(mol\right)\)

\(CH_4+2O_2--t^o->CO_2+2H_2O\)

Ta có: \(n_{O_2}=2.n_{CH_4}=0,1\left(mol\right)\Rightarrow V_{O_2\left(đktc\right)}=2,24\left(l\right)\)

\(M_A=5.2=10\left(g/mol\right)\)

Do các khí đo ở cùng điều kiện nhiệt độ và áp suất nên tỉ lệ thể tích cũng là tỉ lệ mol

Chọn \(\left\{{}\begin{matrix}n_{H_2}=17,5\left(mol\right)\\n_{N_2}=5\left(mol\right)\end{matrix}\right.\)

Gọi \(n_{N_2\left(p\text{ư}\right)}=a\left(mol\right)\left(0< a< 5\right)\)

PTHH: \(N_2+3H_2\xrightarrow[]{t^o,p,xt}2NH_3\)

             a---->3a---------->2a

Xét tỉ lệ: \(5< \dfrac{17,5}{3}\Rightarrow\) Hiệu suất phản ứng tính theo N₂

Ta có: \(n_A=5+17,5+2a-a-3a=22,5-2a\left(mol\right)\)

Theo ĐLBTKL: \(m_A=5.28+17,5.2=175\left(g\right)\)

\(\Rightarrow M_A=\dfrac{175}{22,5-2a}=10\Leftrightarrow a=2,5\left(TM\right)\)

\(\Rightarrow H=\dfrac{2,5}{5}.100\%=50\%\)