Question

при спалюванні 1,84 г вуглеводню утворилося 3,136 л вуглекислого газу і 1, 44 г води. Молярна маса води 92 г/ моль. Знайти формулу сполуки.. ... ТЕРМІНОВО ХІМІЯ КР​

Answer 1

Відповідь:

Пояснення:

Визначаємо мольне співвідношення речовин, що вступили в реакцію

n(CxHy) = 1,84 / 92г/моль= 0,02 моль

n(CO2) = 3,136 / 22,4 л/моль = 0,14 моль

n(H2O) = 1,44 / 18г/моль = 0,08 моль

Приводимо до цілих чисел

1 моль : 7 моль : 4 моль (*2) - це індекси

бруто-формула буде С7Н8 - Mr - 12* 7 + 1*8 = 92 г/моль

Answer 2

To find the formula of the compound formed when 1.84 g of hydrocarbon is burned, we need to use the given information about the products formed during the combustion reaction.

According to the information provided, 1.84 g of hydrocarbon produces 3.136 L of carbon dioxide (CO2) and 1.44 g of water (H2O). The molar mass of water is given as 92 g/mol.

To find the formula of the compound, we can follow these steps:

1. Convert the mass of carbon dioxide to moles: - The molar mass of carbon dioxide (CO2) is 44 g/mol. - Using the formula: moles = mass / molar mass, we can calculate the number of moles of carbon dioxide produced. - moles of CO2 = 3.136 L * (1 mol/22.4 L) * (44 g/mol) = 2.98 g

2. Convert the mass of water to moles: - Using the same formula: moles = mass / molar mass, we can calculate the number of moles of water produced. - moles of H2O = 1.44 g / 92 g/mol = 0.0157 mol

3. Determine the ratio of moles of carbon dioxide to moles of water: - Divide the moles of carbon dioxide by the moles of water to find the ratio. - ratio = moles of CO2 / moles of H2O = 2.98 g / 0.0157 mol = 189.8

4. Simplify the ratio to obtain the empirical formula: - Divide the ratio by the greatest common divisor to simplify it. - The greatest common divisor of 189.8 is 1. - Therefore, the empirical formula is CO2H2O.

Based on the given information, the empirical formula of the compound formed when 1.84 g of hydrocarbon is burned is CO2H2O.

Please note that this calculation assumes complete combustion of the hydrocarbon and does not take into account any other possible products or reactions that may occur.