a, Ta có: \(n_{C_2H_2}=\dfrac{8,96}{22,4}=0,4\left(mol\right)\)

PT: \(2C_2H_2+5O_2\underrightarrow{^{t^o}}4CO_2+2H_2O\)

\(n_{O_2}=\dfrac{5}{2}n_{C_2H_2}=0,5\left(mol\right)\Rightarrow V_{O_2}=0,5.22,4=11,2\left(l\right)\)

b, \(V_{kk}=\dfrac{V_{O_2}}{20\%}=56\left(l\right)\)

c, - Hiện tượng: Br₂ nhạt màu dần.

PT: \(C_2H_2+2Br_2\rightarrow C_2H_2Br_4\)

\(n_{C_2H_4}=\dfrac{8,96}{22,4}=0,4\left(mol\right)\\ PTHH:C_2H_4+3O_2\rightarrow\left(t^o\right)2CO_2+2H_2O\\ n_{O_2}=3.0,4=1,2\left(mol\right);n_{CO_2}=0,4.2=0,8\left(mol\right)\\ a,V_{O_2\left(đktc\right)}=22,4.1,2=26,88\left(l\right)\\ b,V_{kk\left(đktc\right)}=\dfrac{100}{20}.26,88=134,4\left(l\right)\\ c,CO_2+Ca\left(OH\right)_2\rightarrow CaCO_3\downarrow\left(trắng\right)+H_2O\\ n_{CaCO_3}=n_{CO_2}=0,8\left(mol\right)\\ m_{kết.tủa}=m_{CaCO_3}=100.0,8=80\left(g\right)\)

\(2C_2H_2+5O_2\rightarrow\left(t^o\right)4CO_2+2H_2O\\ n_{C_2H_2}=\dfrac{5,6}{22,4}=0,25\left(mol\right)\\ n_{CO_2}=2.0,25=0,5\left(mol\right)\\ a,V_{CO_2\left(đktc\right)}=0,5.22,4=11,2\left(l\right)\\ b,n_{O_2}=\dfrac{5}{2}.0,25=0,625\left(mol\right)\\ V_{O_2\left(đktc\right)}=0,625.22,4=14\left(l\right)\\ V_{kk\left(đkct\right)}=\dfrac{100}{20}.14=70\left(lít\right)\)

\(n_{C_2H_2}=\dfrac{8,96}{22,4}=0,4mol\)

\(2C_2H_2+5O_2\rightarrow\left(t^o\right)4CO_2+2H_2O\)

0,4             1                 0,8                  ( mol )

\(V_{CO_2}=0,8.22,4=17,92l\)

\(V_{kk}=V_{O_2}.5=1.22,4.5=112l\)

\(n_{C_2H_2}=\dfrac{8,96}{22,4}=0,4\left(mol\right)\\ PTHH:2C_2H_2+5O_2\underrightarrow{t^o}4CO_2+2H_2O\\ Mol:0,4\rightarrow1\rightarrow0,8\\ \rightarrow\left\{{}\begin{matrix}V_{CO_2}=0,8.22,4=17,92\left(l\right)\\V_{kk}=1.5.22,4=112\left(l\right)\end{matrix}\right.\)

a) \(n_{C_2H_4}=\dfrac{6,72}{22,4}=0,3\left(mol\right)\)

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

           0,3---->0,9------>0,6

\(\Rightarrow V_{kk}=\dfrac{0,9.22,4}{20\%}=100,8\left(l\right)\)

b) \(V_{CO_2}=0,6.22,4=13,44\left(l\right)\)

c) \(n_P=\dfrac{15,5}{31}=0,5\left(mol\right)\)

PTHH: \(4P+5O_2\xrightarrow[]{t^o}2P_2O_5\)

Xét tỉ lệ: \(\dfrac{0,5}{4}< \dfrac{0,9}{5}\Rightarrow\) O₂ dư, P cháy hết

\(n_{C_2H_4}=\dfrac{8,96}{22,4}=0,4\left(mol\right)\)

C₂H₄ + 3O₂ ----t^o---> 2CO₂ + 2H₂O

0,4        1,2                                0,8

\(m_{H_2O}=0,8.18=14,4\left(g\right)\)

\(V_{kk}=5V_{O_2}=5.1,2.22,4=134,4\left(l\right)\)

\(n_{C_2H_2}=\dfrac{8.96}{22.4}=0.4\left(mol\right)\)

\(C_2H_2+\dfrac{5}{2}O_2\underrightarrow{^{t^0}}2CO_2+H_2O\)

\(0.4.........1\)

\(V_{O_2}=1\cdot22.4=22.4\left(l\right)\)

\(V_{kk}=5V_{O_2}=5\cdot22.4=112\left(l\right)\)