• Matéria: Química
  • Autor: 2020320145
  • Perguntado 4 anos atrás

O balanceamento d) C4H8O(v) + O2(g) → CO2(g) + H2O(v)

Respostas

respondido por: viniciusjosedas
0

Resposta:

C4H8O(v) +  11/2 O2(g) → 4 CO2(g) + 4 H2O(v)

respondido por: Anônimo
1

Resposta:

2C_4H_8O+11O_2=8CO_2+8H_2O\\\\C \ balanceada\\\\H \  balanceada\\\\O \  balanceada\\\\u2C_4H_8O+v11O_2=w8CO_2+t8H_2O\\\\\mathrm{Para\:balancear\:C:}\:8u=8w\\\\\mathrm{Para\:balancear\:H:}\:16u=16t\\\\\mathrm{Para\:balancear\:O:}\:2u+22v=16w+8t\\\\\begin{bmatrix}8u=8w\\ 16u=16t\\ 2u+22v=16w+8t\end{bmatrix}\\\\8u=8w\\\\\frac{8u}{8}=\frac{8w}{8}\\\\u=w\\\\\begin{bmatrix}16w=16t\\ 2w+22v=16w+8t\end{bmatrix}\\\\16w=16t\\\\\frac{16w}{16}=\frac{16t}{16}\\\\

w=t\\\\\begin{bmatrix}2t+22v=16t+8t\end{bmatrix}\\\\2t+22v=16t+8t\\\\16t+8t=24t\\\\2t+22v=24t\\\\\begin{bmatrix}2t+22v=24t\end{bmatrix}\\\\2t+22v-22v=24t-22v\\\\2t=24t-22v\\\\2t-24t=24t-22v-24t\\\\-22t=-22v\\\\\frac{-22t}{-22}=\frac{-22v}{-22}\\\\\frac{-22t}{-22}\\\\=t\\\\\frac{-22v}{-22}\\\\=v\\\\w=v\\\\u=v\\\\u=v,\:w=v,\:t=v\\\\u=1,\:w=1,\:t=1,\:v=1\\\\\mathrm{Portanto\:a\:reacao\:quimica\:balanceada\:e:}\\\\2C_4H_8O+11O_2=8CO_2+8H_2O

Perguntas similares