Question

Опредилите обьем углекислого газа при сжигании 20 г бутана

Answer 1

 20 г                                             Х л
 2С4Н10       +       13О2    ->    8CO2      +    10H2O
n = 2 моль                               n = 8 моль
М = 58 г/моль                        Vm = 22,4 л/моль
m = 116 г                                V = 179,2 л
20 г С4Н10   -   Х л СО2
116 г С4Н10  -   179,2 л СО2
V(CO2) = 20 * 179,2 / 116 = 31 л

Answer 2

Calculating the Volume of Carbon Dioxide Produced from Burning 20g of Butane

To calculate the volume of carbon dioxide produced from burning 20g of butane, we can use the stoichiometry of the combustion reaction of butane.

Step 1: Write the Balanced Chemical Equation for the Combustion of Butane

The balanced chemical equation for the combustion of butane is:

C4H10 + 6.5O2 → 4CO2 + 5H2O

Step 2: Calculate the Moles of Butane

To calculate the moles of butane, we use the formula:

moles = mass / molar mass

The molar mass of butane (C4H10) is 58.12 g/mol.

moles of butane = 20g / 58.12 g/mol = 0.344 moles

Step 3: Use Stoichiometry to Calculate the Moles of Carbon Dioxide

From the balanced chemical equation, we can see that 1 mole of butane produces 4 moles of carbon dioxide.

moles of carbon dioxide = 0.344 moles * 4 = 1.376 moles

Step 4: Convert Moles to Volume of Carbon Dioxide

The volume of 1 mole of any gas at standard temperature and pressure (STP) is 22.4 liters.

volume of carbon dioxide = 1.376 moles * 22.4 L/mol = 30.82 liters

Conclusion: The volume of carbon dioxide produced from burning 20g of butane is approximately 30.82 liters.