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\(n_{C_2H_2}=\dfrac{2,8}{22,4}=0,125\left(mol\right)\\ a,2C_2H_2+5O_2\rightarrow\left(t^o\right)4CO_2+2H_2O\\ b,n_{CO_2}=0,125.2=0,25\left(mol\right)\\ m_{CO_2}=0,25.44=11\left(g\right)\\ c,n_{O_2}=\dfrac{5}{2}.0,125=0,3125\left(mol\right)\\ V_{O_2\left(đktc\right)}=0,3125.22,4=7\left(l\right)\\ \Rightarrow V_{kk\left(đktc\right)}=\dfrac{100}{20}.7=35\left(l\right)\)
\(2C_2H_2+5O_2\rightarrow\left(t^o\right)4CO_2+2H_2O\\ n_{C_2H_2}=\dfrac{5,6}{22,4}=0,25\left(mol\right)\\ n_{CO_2}=2.0,25=0,5\left(mol\right)\\ a,V_{CO_2\left(đktc\right)}=0,5.22,4=11,2\left(l\right)\\ b,n_{O_2}=\dfrac{5}{2}.0,25=0,625\left(mol\right)\\ V_{O_2\left(đktc\right)}=0,625.22,4=14\left(l\right)\\ V_{kk\left(đkct\right)}=\dfrac{100}{20}.14=70\left(lít\right)\)
\(n_{C_2H_2}=\dfrac{5,6}{22,4}=0,25mol\)
\(C_2H_2+\dfrac{5}{2}O_2\underrightarrow{t^o}2CO_2+H_2O\)
0,25 0,625
\(V_{O_2}=0,625\cdot22,4=14l\)
\(V_{kk}=5V_{O_2}=5\cdot14=70l\)
\(n_{C_2H_2}=\dfrac{5,6}{22,4}=0,25mol\)
\(2C_2H_2+5O_2\rightarrow\left(t^o\right)4CO_2+2H_2O\)
0,25 0,625 ( mol )
\(V_{O_2}=0,625.22,4=14l\)
\(V_{kk}=V_{O_2}.5=14.5=70l\)
\(n_{C_2H_2}=\dfrac{8,96}{22,4}=0,4mol\)
\(2C_2H_2+5O_2\rightarrow\left(t^o\right)4CO_2+2H_2O\)
0,4 1 0,8 ( mol )
\(V_{CO_2}=0,8.22,4=17,92l\)
\(V_{kk}=V_{O_2}.5=1.22,4.5=112l\)
\(n_{C_2H_2}=\dfrac{8,96}{22,4}=0,4\left(mol\right)\\ PTHH:2C_2H_2+5O_2\underrightarrow{t^o}4CO_2+2H_2O\\ Mol:0,4\rightarrow1\rightarrow0,8\\ \rightarrow\left\{{}\begin{matrix}V_{CO_2}=0,8.22,4=17,92\left(l\right)\\V_{kk}=1.5.22,4=112\left(l\right)\end{matrix}\right.\)
\(n_{C_2H_4}=\dfrac{8,96}{22,4}=0,4\left(mol\right)\\ PTHH:C_2H_4+3O_2\rightarrow\left(t^o\right)2CO_2+2H_2O\\ n_{O_2}=3.0,4=1,2\left(mol\right);n_{CO_2}=0,4.2=0,8\left(mol\right)\\ a,V_{O_2\left(đktc\right)}=22,4.1,2=26,88\left(l\right)\\ b,V_{kk\left(đktc\right)}=\dfrac{100}{20}.26,88=134,4\left(l\right)\\ c,CO_2+Ca\left(OH\right)_2\rightarrow CaCO_3\downarrow\left(trắng\right)+H_2O\\ n_{CaCO_3}=n_{CO_2}=0,8\left(mol\right)\\ m_{kết.tủa}=m_{CaCO_3}=100.0,8=80\left(g\right)\)
\(n_{C_2H_2}=\dfrac{8.96}{22.4}=0.4\left(mol\right)\)
\(C_2H_2+\dfrac{5}{2}O_2\underrightarrow{^{t^0}}2CO_2+H_2O\)
\(0.4.........1\)
\(V_{O_2}=1\cdot22.4=22.4\left(l\right)\)
\(V_{kk}=5V_{O_2}=5\cdot22.4=112\left(l\right)\)
a, \(n_{C_2H_4}=\dfrac{1,12}{22,4}=0,05\left(mol\right)\)
PT: \(C_2H_4+3O_2\underrightarrow{t^o}2CO_2+2H_2O\)
\(CO_2+Ca\left(OH\right)_2\rightarrow CaCO_{3\downarrow}+H_2O\)
Theo PT: \(n_{CaCO_3}=n_{CO_2}=2n_{C_2H_4}=0,1\left(mol\right)\)
\(\Rightarrow m_{CaCO_3}=0,1.100=10\left(g\right)\)
b, Theo PT: \(n_{O_2}=3n_{C_2H_4}=0,15\left(mol\right)\Rightarrow V_{O_2}=0,15.22,4=3,36\left(l\right)\)
\(\Rightarrow V_{kk}=\dfrac{V_{O_2}}{20\%}=16,8\left(l\right)\)
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: \(C_2H_2+2Br_2\rightarrow C_2H_2Br_4\)