Question

У результаті спалювання 60 г ацетиленового вуглеводню одержали CO2 і H2O, загальна маса яких 252 г. Обчислити об'єм кисню, який витратили на реакцію.

Answer 1

Пишем уравнение горения  и составляем соотношение масс алкина и продуктов

60 г…………………………………..    252 г

СхН2х-2     +0,5(3х-1) О2 =    хСО2 + (х-1)Н2О

(12х+2х-2) г  ………………………(44х+(х-1)18) г

60      -     252

(14х-2)  - (62х-18 )

252* (14х-2) =  60*(62х-18 )

3528х-504 =3720х-1080

576=192х

х=3, С3Н4

С3Н4 + 4О2 + 3СО2+2Н2О

n(С3Н4)= m/M=60/40=1,5 моль

n(О2)=4* n(С3Н4)=1,5*4=6 моль

V(O2)= n(О2)*Vm=6*22.4=134.4 л

Answer 2

Calculation of the Oxygen Volume Used in the Reaction

To calculate the volume of oxygen used in the reaction, we need to determine the stoichiometry of the reaction and use the ideal gas law.

The balanced chemical equation for the combustion of acetylene (C2H2) is:

2 C2H2 + 5 O2 -> 4 CO2 + 2 H2O

From the equation, we can see that 2 moles of acetylene react with 5 moles of oxygen to produce 4 moles of carbon dioxide and 2 moles of water.

Given that the total mass of carbon dioxide and water produced is 252 g, we can calculate the number of moles of carbon dioxide and water produced:

- The molar mass of carbon dioxide (CO2) is 44 g/mol, so the number of moles of CO2 produced is 252 g / 44 g/mol = 5.727 moles. - The molar mass of water (H2O) is 18 g/mol, so the number of moles of H2O produced is 252 g / 18 g/mol = 14 moles.

Since the stoichiometry of the reaction tells us that 5 moles of oxygen react with 2 moles of acetylene, we can set up a proportion to find the number of moles of oxygen used:

(5 moles O2) / (2 moles C2H2) = (x moles O2) / (60 g C2H2)

Solving for x, we find that x = (5/2) * (60 g C2H2) / (26 g/mol C2H2) = 57.69 moles of oxygen.

Now, to calculate the volume of oxygen used, we can use the ideal gas law, which states that:

PV = nRT

Where: - P is the pressure of the gas (assume standard pressure, 1 atm) - V is the volume of the gas - n is the number of moles of the gas - R is the ideal gas constant (0.0821 L·atm/(mol·K)) - T is the temperature in Kelvin (assume room temperature, around 298 K)

Rearranging the equation, we can solve for V:

V = (nRT) / P

Substituting the values, we get:

V = (57.69 moles) * (0.0821 L·atm/(mol·K)) * (298 K) / (1 atm) = 1392.7 L

Therefore, the volume of oxygen used in the reaction is approximately 1392.7 liters.

[[1]]