Question

Answer 1

Theo gt ta có: $n_{C_2H_4}=0,5(mol)$

$C_2H_4+3O_2\rightarrow 2CO_2+2H_2O$

a, Ta có: $n_{O_2}=1,5(mol)\Rightarrow V_{O_2}=33,6(l)$

b, Ta có: $n_{CO_2}=1(mol)\Rightarrow m_{CO_2}=44(g)$

$n_{H_2O}=1(mol)\Rightarrow m_{H_2O}=18(g)$

 

Answer 2

\(n_{C_2H_4}=\dfrac{11.2}{22.4}=0.5\left(mol\right)\)

\(C_2H_4+3O_2\underrightarrow{^{t^0}}2CO_2+2H_2O\)

\(0.5.......1.5...........1...........1\)

\(V_{O_2}=1.5\cdot22.4=33.6\left(l\right)\)

\(m_{CO_2}=1\cdot44=44\left(g\right)\)

\(m_{H_2O}=1\cdot18=18\left(g\right)\)