a) PTHH:
\(C_2H_4+3O_2\xrightarrow[]{t^o}2CO_2+2H_2O\) (1)

\(2C_2H_2+5O_2\xrightarrow[]{t^o}4CO_2+2H_2O\) (2)

\(CO_2+Ba\left(OH\right)_2\rightarrow BaCO_3\downarrow+H_2O\) (3)

\(C_2H_4+Br_2\rightarrow C_2H_4Br_2\) (4)

\(C_2H_2+2Br_2\rightarrow C_2H_2Br_4\) (5)

b) Gọi \(\left\{{}\begin{matrix}n_{C_2H_4}=x\left(mol\right)\\n_{C_2H_2}=y\left(mol\right)\end{matrix}\right.\)

=> \(x+y=\dfrac{8,4}{22,4}=0,375\left(I\right)\)

Theo PT (1), (2): \(n_{O_2}=3n_{C_2H_4}+\dfrac{5}{2}n_{C_2H_2}\)

=> \(3x+2,5y=\dfrac{22,4}{22,4}=1\left(II\right)\)

Từ (I), (II) => x = 0,125; y = 0,25

BTNT C: \(n_{BaCO_3}=n_{CO_2}=2n_{C_2H_2}+2n_{C_2H_4}=0,75\left(g\right)\)

=> \(a=m_{BaCO_3}=0,75.197=147,75\left(g\right)\)

Ta có: \(b=m_{tăng}=m_{C_2H_2}+m_{C_2H_4}=0,125.28+0,25.26=10\left(g\right)\)

a) Gọi \(\left\{{}\begin{matrix}n_{Al}=a\left(mol\right)\\n_{Fe}=b\left(mol\right)\end{matrix}\right.\)

PTHH: 
2Al + 3H₂SO₄ ---> Al₂(SO₄)₃ + 3H₂ 

a------------------------------------->1,5a

Fe + H₂SO₄ ---> FeSO₄ + H₂ 

b------------------------------>b

b) Ta có hệ PT: \(\left\{{}\begin{matrix}27a+56b=0,83\\1,5a+b=\dfrac{0,56}{22,4}=0,025\end{matrix}\right.\Leftrightarrow a=b=0,01\)

=> \(\left\{{}\begin{matrix}\%m_{Al}=\dfrac{0,01.27}{0,83}.100\%=32,53\%\\\%m_{Fe}=100\%-32,53\%=67,47\%\end{matrix}\right.\)

$\rm H_2S: S^{2-}; H^{+1}$
$\rm NO: N^{+2}; O^{-2}$
$\rm N_2O_5: N^{+5}; O^{-2}$
$\rm H_2SO_4: H^{+1}; S^{+6}; O^{-2}$
$\rm Ba(OH)_2: Ba^{+2}; O^{-2}; H^{+1}$
$\rm K_2SO_4: K^{+1}; S^{+6}; O^{-2}$
$\rm K_2Cr_2O_7: K^{+1}; Cr^{+6}; O^{-2}$

\(ZnS+O->2SO_2+ZnO\) câu này xem lại đề 

\(Al_2O_3+6HCl->2AlCl_3+3H_2O\)

\(2KMnO_4->K_2MnO_4+O_2+MnO_2\)

\(4Na+O_2->2Na_2O\)

\(2Na+2H_2O->2NaOH+H_2\)

\(Cu\left(OH\right)_2->H_2O+CuO\)

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

\(2Na+Cl_2->2NaCl\)

\(m_{CaO\left(TT\right)}=112.90\%=100,8\left(kg\right)\\ m_{Ca\left(OH\right)_2}=\dfrac{100,8.74}{56}=133,2\left(kg\right)\)

cm2 là sao ạ?

Thể tích phải là 36cm³ chứ em?

a. \(n_{Ba\left(OH\right)_2}=\dfrac{200.17,1\%}{171}=0,2\left(mol\right)\)

PTHH : Ba(OH)₂ + SO₂ -> BaSO₃ \(\downarrow\) + H₂O

               0,2            0,2         0,2

b. \(m_{BaSO_3}=0.2.217=43,4\left(g\right)\)

c. \(V_{SO_2}=0,2.22,4=4,48\left(l\right)\)

a) Ta có: \(n_{Ba\left(OH\right)_2}=\dfrac{200.17,1\%}{171}=0,2\left(mol\right)\)

PTHH: \(Ba\left(OH\right)_2+SO_2\rightarrow BaSO_3\downarrow+H_2O\)

              0,2-------->0,2------->0,2                         (mol)

b) m_{kết tủa} = \(m_{BaSO_3}=0,2.217=43,4\left(g\right)\)

c) \(V_a=V_{SO_2}=0,2.22,4=4,48\left(l\right)\)

theo định luật bảo toàn năng lượng ta có :

           tổng khối lượng các chất tham gia sẽ bằng với tổng khối lượng sản phẩm

Do đó : M_O⁺           M_H= M sản phẩm

suy ra : 32 g+ M_H=36g

suy ra M_H=36g-32g=4g

Có: \(D_{H_2O}=1g/cm^3\)

=> \(m_{H_2O}=V_{H_2O}.D_{H_2O}=36.1=36\left(g\right)\)

Theo ĐLBTKL: \(m_{O_2}+m_{H_2}=m_{H_2O}\)

=> \(m_{H_2}=m_{H_2O}-m_{O_2}=36-32=4\left(g\right)\)

a) Ta có: \(M_A=20.2=40\left(g/mol\right)\)

Gọi \(\left\{{}\begin{matrix}n_{O_2}=a\left(mol\right)\\n_{O_3}=b\left(mol\right)\end{matrix}\right.\Rightarrow\left\{{}\begin{matrix}m_{O_2}=32a\left(g\right)\\m_{O_3}=48b\left(g\right)\end{matrix}\right.\)

=> \(\left\{{}\begin{matrix}n_A=n_{O_2}+n_{O_3}=a+b\left(mol\right)\\m_A=m_{O_2}+m_{O_3}=32a+48b\left(g\right)\end{matrix}\right.\)

=> \(M_A=\dfrac{m_A}{n_A}=\dfrac{32a+48b}{a+b}=40\left(g/mol\right)\)

=> 32a + 48b = 40a + 40b

=> 8b = 8a

=> a = b

Ta có %V cũng là %n

=> \(\left\{{}\begin{matrix}\%V_{O_2}=\%n_{O_2}=\dfrac{a}{a+b}.100\%=\dfrac{a}{a+a}.100\%=\dfrac{1}{2}.100\%=50\%\\\%V_{O_3}=100\%-50\%=50\%\end{matrix}\right.\)

b) 1 mol khí có \(\left\{{}\begin{matrix}n_{O_2}=\dfrac{1.50}{100}=0,5\left(mol\right)\\n_{O_3}=1-0,5=0,5\left(mol\right)\end{matrix}\right.\)

PTHH: \(CH_4+2O_2\xrightarrow[]{t^o}CO_2+2H_2O\left(1\right)\)

            \(3CH_4+4O_3\xrightarrow[]{t^o}3CO_2+6H_2O\left(2\right)\)

Theo PTHH (1), (2): \(n_{CH_4}=\dfrac{3}{4}.n_{O_3}+\dfrac{1}{2}.n_{O_2}=\dfrac{3}{4}.0,5+\dfrac{1}{2}.0,5=0,625\left(mol\right)\)

=> \(V_{CH_4}=0,625.22,4=14\left(l\right)\)