Question

При спалюванні 40мл газу було витрачено 340мл кисню і одержано 240мл карбон (IV) оксиду та 200мл водяної пари. Що це за сполука? помогите, пожалуйста ​

Answer 1

цей газ С2Н4 - етен (етилен)

підставляємо дані, як молі(коефіцієнти) в рівняння і знаходимо формулу газу:

СₐНₓ + 3О₂ = 2СО₂ +2Н₂О

С₂Н₄ + 3О₂ = 2СО₂ +2Н₂О

Answer 2

Chemical Reaction Analysis

The given information describes a chemical reaction involving the combustion of a gas. Let's analyze the reaction based on the provided data.

The reaction can be represented as follows: CnHm + (n + 1/4m)O2 → nCO2 + (m/2)H2O

Where: - CnHm represents the hydrocarbon gas - (n + 1/4m)O2 represents the oxygen consumed - nCO2 represents the carbon dioxide produced - (m/2)H2O represents the water vapor produced

Calculation of the Chemical Equation

Based on the given data: - Gas consumed: 40ml - Oxygen consumed: 340ml - Carbon (IV) oxide produced: 240ml - Water vapor produced: 200ml

From the given information, we can calculate the molar ratio of the reactants and products.

The molar ratio of the gas to oxygen can be calculated as follows: - 40ml gas consumed corresponds to 340ml oxygen consumed

The molar ratio of carbon (IV) oxide to water vapor can be calculated as follows: - 240ml carbon (IV) oxide produced corresponds to 200ml water vapor produced

Identification of the Compound

Based on the provided information, the compound formed from the given reaction is carbon dioxide (CO2). This is evident from the production of 240ml of carbon (IV) oxide and 200ml of water vapor.

The balanced chemical equation for the reaction is: CnHm + (n + 1/4m)O2 → nCO2 + (m/2)H2O

Therefore, the compound formed from the given reaction is carbon dioxide (CO2).

I hope this helps! If you have further questions or need additional assistance, feel free to ask.