a) \(n_{Br_2}=0,1.2=0,2\left(mol\right)\)

PTHH: \(CH_2=CH_2+Br_2\rightarrow CH_2Br-CH_2Br\)

                  0,2<-------0,2

\(\Rightarrow V_{C_2H_4}=0,2.22,4=4,48\left(l\right)\)

b) PTHH: \(CH\equiv CH+2Br_2\rightarrow CHBr_2-CHBr_2\)

                   0,1<--------0,2

\(\Rightarrow n_{C_2H_4}-n_{C_2H_2}=0,1\left(mol\right)\)

Sửa đề: 5,04 gam -> 5,4 gam

a) 

Theo ĐLBTNT: \(\left\{{}\begin{matrix}n_H=2n_{H_2O}=2.\dfrac{5,4}{18}=0,6\left(mol\right)\\n_C=n_{CO_2}=\dfrac{6,72}{22,4}=0,3\left(mol\right)\\n_O=\dfrac{9-0,3.12-0,6}{16}=0,3\left(mol\right)\end{matrix}\right.\)

Vậy X chứa C, H, O

b) Đặt CTPT của X là C_xH_yO_z

\(\Rightarrow n_X=\dfrac{9}{180}=0,05\left(mol\right)\Rightarrow\left\{{}\begin{matrix}x=\dfrac{0,3}{0,06}=6\\y=\dfrac{0,6}{0,05}=12\\z=\dfrac{0,3}{0,05}=6\end{matrix}\right.\)

Vậy X là C₆H₁₂O₆

a) Thu nhiệt

b) Tỏa nhiệt

c) Thu nhiệt

d) Tỏa nhiệt

e) Thu nhiệt

f) Tỏa nhiệt

g) Tỏa nhiệt

i) Thu nhiệt

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

PTHH: \(4Al+3O_2\xrightarrow[]{t^o}2Al_2O_3\)

          0,2--->0,15

b) \(V_{O_2}=0,15.22,4=3,36\left(l\right)\)

c) PTHH: \(2KMnO_4\xrightarrow[]{t^o}K_2MnO_4+MnO_2+O_2\)

                  0,3<-----------------------------------0,15

\(\Rightarrow m_{KMnO_4}=0,3.158=47,4\left(g\right)\)

a) \(2Fe+3Cl_2\xrightarrow[]{t^o}2FeCl_3\)

b) \(Fe+S\xrightarrow[]{t^o}FeS\)

c) \(2NaHCO_3\xrightarrow[]{t^o}Na_2CO_3+CO_2+H_2O\)

1) Si, P, N O, F

2) Rb, K, Ca, Mg, Al

a) \(\left\{{}\begin{matrix}n_{Fe}=\dfrac{11,2}{56}=0,2\left(mol\right)\\n_{HCl}=\dfrac{10,95}{36,5}=0,3\left(mol\right)\end{matrix}\right.\)

PTHH: \(Fe+2HCl\rightarrow FeCl_2+H_2\)

Xét tỉ lệ: \(0,2>\dfrac{0,3}{2}\Rightarrow\) Fe dư

Theo PTHH: \(n_{Fe\left(p\text{ư}\right)}=n_{H_2}=\dfrac{1}{2}n_{HCl}=0,15\left(mol\right)\)

\(\Rightarrow m_{Fe\left(d\text{ư}\right)}=\left(0,2-0,15\right).56=2,8\left(g\right)\)

c) \(V_{H_2}=0,15.22,4=3,36\left(l\right)\)

\(n_{HCl}=0,1.1=0,1\left(mol\right);n_{H_2SO_4}=0,1.1=0,1\left(mol\right)\)

PTHH: 

`NaOH + HCl -> NaCl + H_2O`

`2NaOH + H_2SO_4 -> Na_2SO_4 + 2H_2O`

Theo PT: `n_{NaOH} = 2n_{H_2SO_4} + n_{HCl} = 0,3 (mol)`

`=> V_{ddNaOH} = (0,3)/(1) = 0,3(l)`

Ca(HCO₃)₂, NaHCO₃, KHCO₃ => Chọn B

a) \(n_{H_2SO_4}=\dfrac{24,5}{98}=0,25\left(mol\right)\)

PTHH: Zn + H₂SO₄ ---> ZnSO₄ + H₂

           0,25<---0,25----------------->0,25

=> m_Zn = 0,25.65 = 16,25 (g)

b) \(V_{H_2}=0,25.22,4=5,6\left(l\right)\)