Kiểm tra chương IV Oxi-Không khí: Đề 2

a, PT: \(4Al+3O_2\underrightarrow{t^o}2Al_2O_3\)

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

Theo PT: \(n_{Al_2O_3}=\dfrac{1}{2}n_{Al}=0,1\left(mol\right)\Rightarrow m_{Al_2O_3}=0,1.102=10,2\left(g\right)\)

c, Theo PT: \(n_{O_2}=\dfrac{3}{4}n_{Al}=0,15\left(mol\right)\Rightarrow V_{O_2}=0,15.22,4=3,36\left(l\right)\)

\(\Rightarrow V_{kk}=\dfrac{V_{O_2}}{20\%}=16,8\left(l\right)\)

d, PT: \(2KClO_3\underrightarrow{t^o}2KCl+3O_2\)

Theo PT: \(n_{KClO_3}=\dfrac{2}{3}n_{O_2}=0,1\left(mol\right)\Rightarrow m_{KClO_3}=0,1.122,5=12,25\left(g\right)\)

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--->0,1

b) \(m_{Al_2O_3}=0,1.102=10,2\left(g\right)\)

c) \(V_{kk}=\dfrac{0,15.22,4}{20\%}=16,8\left(l\right)\)

d) \(2KClO_3\xrightarrow[]{t^o}2KCl+3O_2\)

        0,1<-----------------0,15

\(\Rightarrow m_{KClO_3}=0,1.122,5=12,25\left(g\right)\)

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

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

`2K + 2H_2 O -> 2KOH + H_2↑`

`0,1`     `0,1`                           `0,05`      `(mol)`

`n_K = [ 3,9 ] / 39 = 0,1 (mol)`

`a) V_[H_2] = 0,05 . 22,4 = 1,12 (l)`

`b)m_[H_2 O] = 0,1 . 18 = 1,8 (g)`

4Al+3O2-to>2Al2O3

4K+O2-to>2K2O

S+O2-to>SO2

C2H4+3O2-to>2CO2+2H2O

4Al + 3O2 -> (t°) 2Al2O3

4K + O2 -> 2K2O

S + O2 -> (t°) SO2

C2H4 + 3O2 -> (t°) 2CO2 + 2H2O

4Al + 3O₂ ⇒ 2Al₂O₃

4K + O₂ ⇒ 2K₂O

S + O₂ ⇒ SO₂

C₂H₄ + 3O₂ ⇒ 2CO₂ + 2H₂O (hình như đề câu này bạn viết sai)

C + O2 ---> CO2

C + O2 --> CO2

C+O2-to>CO2

2CO+O2-to>2CO2

\(n_P=\dfrac{6,2}{31}=0,2mol\)

\(n_{O_2}=\dfrac{2,479}{22,4}\approx0,11mol\)

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

Bđ    0,2      0,11

Pư  0,088    0,11    0,044

Kt    0,112     0        0,044

a)\(m=m_{P_2O_5}=0,044\cdot142=6,248g\)

b)\(m_{dư}=m_P=0,112\cdot31=3,472g\)

\(a.Mg+2HCl\rightarrow MgCl_2+H_2\\ 2Al+6HCl\rightarrow2AlCl_3+3H_2\\ b.Đặt:\left\{{}\begin{matrix}n_{Mg}=x\left(mol\right)\\n_{Al}=y\left(mol\right)\end{matrix}\right.\\ n_{H_2}=\dfrac{18,48}{22,4}=0,825\left(mol\right)\\ Tacó:\left\{{}\begin{matrix}24x+27y=17,1\\x+\dfrac{3}{2}y=0.825\end{matrix}\right.\\ \Rightarrow\left\{{}\begin{matrix}x=0,375\\y=0,3\end{matrix}\right.\\ \Rightarrow\%m_{Mg}=\dfrac{0,375.24}{17,1}.100=52,63\%\\ \%m_{Al}=47,37\%\)

V(O2)= 20%.Vkk=20%. 44,8= 8,92(l) => nO2=0,4(mol)

nC3H8= 1,344/22,4= 0,06(mol)

PTHH: C3H8 +  5 O2 -to-> 3 CO2 + 4 H2O

Ta có: 0,06/1 < 0,4/5

=> O2 dư, C3H8 hết, tính theo nC3H8

=> nO2(p.ứ)= 0,06.5=0,3(mol)=> nO2(dư)=0,4-0,3=0,1(mol)

=> V(O2,dư)=0,1.22,4=2,24(l)

b) nCO2=3.0,06=0,18(mol)

=>mCO2=0,18 . 44=7,92(g)

nH2O=0,06.4=0,24(mol)

=>mH2O=0,24.18=4,32g)

Chúc em học tốt!

â) 4P +5 O2 -to-> 2 P2O5

b) nO2=5,6/22,4=0,25(mol)

=>nP=0,25 x 4/5 =0,2(mol)

=>mP=0,2.31=6,2(g)

c) nP=15,5/31=0,5(mol)

Ta có: 0,5/4 > 0,25/5

-> P dư, O2 hết, tính theo nO2.

=> nP2O5=2/5. 0,25=0,1(mol

=>mP2O5=142.0,1=14,2(g)

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

\(n_{O_2}=0,25\left(mol\right)\)

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

0,2 ←0,25

\(m_P=0,2.31=6,2\left(g\right)\)