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a.\(n_{CH_4}=\dfrac{V_{CH_4}}{22,4}=\dfrac{11,2}{22,4}=0,5mol\)
\(CH_4+2O_2\rightarrow\left(t^o\right)CO_2+2H_2O\)
0,5 1 0,5 ( mol )
\(V_{kk}=V_{O_2}.5=\left(1.22,4\right).5=112l\)
b.\(n_{NaOH}=0,5.0,5=0,25mol\)
\(NaOH+CO_2\rightarrow NaHCO_3\)
0,25 < 0,5 ( mol )
0,25 0,25 ( mol )
\(m_{NaHCO_3}=0,25.84=21g\)
CH4+2O2-to>CO2+2H2O
0,5-----1----------0,5 mol
n CH4=\(\dfrac{11,2}{22,4}\)=0,5 mol
=>Vkk=1.22,4.5=112l
NaOH+CO2->NaHCO3
0,25------0,25-------0,25
n NaOH=0,5.0,5=0,25 mol
=>Tạo ra muối axit, CO2 dư
=>m NaHCO3=0,25.84=21g
\(n_{C_2H_4}=\dfrac{8,96}{22,4}=0,4\left(mol\right)\\ PTHH:C_2H_4+3O_2\rightarrow\left(t^o\right)2CO_2+2H_2O\\ n_{O_2}=3.0,4=1,2\left(mol\right);n_{CO_2}=0,4.2=0,8\left(mol\right)\\ a,V_{O_2\left(đktc\right)}=22,4.1,2=26,88\left(l\right)\\ b,V_{kk\left(đktc\right)}=\dfrac{100}{20}.26,88=134,4\left(l\right)\\ c,CO_2+Ca\left(OH\right)_2\rightarrow CaCO_3\downarrow\left(trắng\right)+H_2O\\ n_{CaCO_3}=n_{CO_2}=0,8\left(mol\right)\\ m_{kết.tủa}=m_{CaCO_3}=100.0,8=80\left(g\right)\)
\(n_{C_2H_4}=\dfrac{8,96}{22,4}=0,4\left(mol\right)\)
C2H4 + 3O2 ----to---> 2CO2 + 2H2O
0,4 1,2 0,8
\(m_{H_2O}=0,8.18=14,4\left(g\right)\)
\(V_{kk}=5V_{O_2}=5.1,2.22,4=134,4\left(l\right)\)
\(n_{C_2H_2}=\dfrac{2,8}{22,4}=0,125\left(mol\right)\\ a,2C_2H_2+5O_2\rightarrow\left(t^o\right)4CO_2+2H_2O\\ b,n_{CO_2}=0,125.2=0,25\left(mol\right)\\ m_{CO_2}=0,25.44=11\left(g\right)\\ c,n_{O_2}=\dfrac{5}{2}.0,125=0,3125\left(mol\right)\\ V_{O_2\left(đktc\right)}=0,3125.22,4=7\left(l\right)\\ \Rightarrow V_{kk\left(đktc\right)}=\dfrac{100}{20}.7=35\left(l\right)\)
\(n_{C_2H_4}=\dfrac{11,2}{22,4}=0,5\left(mol\right)\\ C_2H_4+3O_2\rightarrow\left(t^o\right)2CO_2+2H_2O\\ a,n_{O_2}=3.0,5=1,5\left(mol\right)\\ V_{O_2\left(đktc\right)}=1,5.22,4=33,6\left(l\right)\\ V_{kk\left(đktc\right)}=33,6.5=168\left(l\right)\\ b,n_{CO_2}=n_{H_2O}=2.0,5=1\left(mol\right)\\ m_{CO_2}=44.1=44\left(g\right);m_{H_2O}=18.1=18\left(g\right)\\ c,CO_2+Ca\left(OH\right)_2\rightarrow CaCO_3+H_2O\\ n_{Ca\left(OH\right)_2}=n_{CO_2}=1\left(mol\right)\\ m_{Ca\left(OH\right)_2}=1.74=74\left(g\right)\\ m_{ddCa\left(OH\right)_2}=\dfrac{74.100}{10}=740\left(g\right)\)