Question

Напишите уравнения реакций, которые лежат в основе схемы СН4-А-С2Н6-Б-С4Н10-СО2

Answer 1

CH4+Cl2 -> CH3-Cl + HCl
CH3-Cl + 2Na+ Cl-CH3 -> CH3-CH3 + 2NaCl
CH3-CH3 + Cl2 -> CH3-CH2-Cl + HCl
CH3-CH2-Cl + 2Na + Cl-CH2-CH3 -> CH3-CH2-CH2-CH3 +2NaCl
2C4H10 + 13O2 ->8CO2 + 10H2O

Answer 2

I can help you with writing chemical equations. The scheme you provided shows the transformations of hydrocarbons. Here are the equations of the reactions that underlie the scheme:

- СН4-А-С2Н6: this is a reaction of methane with chlorine, which results in ethane and hydrogen chloride. The equation is $$CH_4 + Cl_2 \rightarrow C_2H_6 + HCl$$ - С2Н6-Б-С4Н10: this is a reaction of ethane with sulfuric acid, which results in butane and water. The equation is $$C_2H_6 + H_2SO_4 \rightarrow C_4H_{10} + H_2O$$ - С4Н10-СО2: this is a reaction of butane with oxygen, which results in carbon dioxide and water. This is a combustion reaction. The equation is $$C_4H_{10} + 13/2 O_2 \rightarrow 4 CO_2 + 5 H_2O$$

You can learn more about how to write and balance chemical equations from [this article](https://externat.foxford.ru/polezno-znat/wiki-himiya-uravneniya-himicheskih-reakcij) or [this website](https://chemequations.com/ru/). I hope this helps.