Question

К смеси углеводорода с аммиаком общ. объемом 300 см3 прибавили избыток кислорода и смесь подожгли. после полного сгорания V смеси стал равен 1250 см3, после конденсации паров воды уменьшился до 550 см3, из которых 100 см3 приходится на азот. установите молекулярную формулу вещества.

Answer 1

Решение в фото(но я не уверен).................

Answer 2

Calculation of the Molecular Formula

To determine the molecular formula of the substance, we need to analyze the given information about the reaction and the volumes of gases involved.

According to the information provided, a mixture of hydrocarbon and ammonia with a total volume of 300 cm3 was mixed with an excess of oxygen and ignited. After complete combustion, the volume of the mixture increased to 1250 cm3. After condensation of water vapor, the volume decreased to 550 cm3, with 100 cm3 of nitrogen present.

Let's break down the information step by step:

1. The initial volume of the mixture (before combustion) is 300 cm3. 2. After complete combustion, the volume of the mixture increases to 1250 cm3. 3. After condensation, the volume decreases to 550 cm3. 4. 100 cm3 of the remaining volume after condensation is nitrogen.

To determine the molecular formula, we need to calculate the moles of each gas involved in the reaction. We can use the ideal gas law to calculate the moles:

PV = nRT

Where: - P is the pressure (assumed constant) - V is the volume - n is the number of moles - R is the ideal gas constant - T is the temperature (assumed constant)

Since the pressure, temperature, and gas constant are assumed constant, we can simplify the equation to:

V = n

Using this equation, we can calculate the number of moles for each gas involved.

Let's calculate the moles of each gas:

1. Moles of nitrogen (N2): - Volume of nitrogen (N2) = 100 cm3 - Moles of nitrogen (N2) = Volume of nitrogen (N2) = 100 cm3

2. Moles of hydrogen (H2): - Volume of hydrogen (H2) = Volume of the mixture after condensation - Volume of nitrogen = 550 cm3 - 100 cm3 = 450 cm3 - Moles of hydrogen (H2) = Volume of hydrogen (H2) = 450 cm3

3. Moles of ammonia (NH3): - Volume of ammonia (NH3) = Volume of the mixture after combustion - Volume of the mixture after condensation = 1250 cm3 - 550 cm3 = 700 cm3 - Moles of ammonia (NH3) = Volume of ammonia (NH3) / 2 (according to the balanced chemical equation) = 700 cm3 / 2 = 350 cm3

Now that we have the moles of each gas, we can determine the molecular formula.

From the calculations above, we have: - Moles of nitrogen (N2) = 100 cm3 - Moles of hydrogen (H2) = 450 cm3 - Moles of ammonia (NH3) = 350 cm3

To find the molecular formula, we need to determine the ratio of the moles of each element in the compound. We can do this by dividing the moles of each element by the smallest number of moles (in this case, the moles of nitrogen).

Dividing the moles of each element by 100 cm3 (the moles of nitrogen), we get: - Moles of nitrogen (N2) = 100 cm3 / 100 cm3 = 1 - Moles of hydrogen (H2) = 450 cm3 / 100 cm3 = 4.5 - Moles of ammonia (NH3) = 350 cm3 / 100 cm3 = 3.5

Now, we need to convert the moles of each element to whole numbers by multiplying them by a common factor. In this case, the common factor is 2, which will give us whole numbers for the moles of each element.

Multiplying the moles of each element by 2, we get: - Moles of nitrogen (N2) = 1 * 2 = 2 - Moles of hydrogen (H2) = 4.5 * 2 = 9 - Moles of ammonia (NH3) = 3.5 * 2 = 7

Therefore, the molecular formula of the substance is N2H4.

Please note that the calculations provided are based on the given information and assumptions.