Question

При спалюванні пентану утворився вуглекислий газ об’ємом 8,96л. Який об’єм вихідної речовин прореагував?

Answer 1

Відповідь:

Дано:                                  СИ                                ₃₀₆ ₄ кДж

V(O₂)=96 л.               4NH₃ + 3O₂ = 2N₂ + 6H₂0 +6153,2 кДж - метод            

                                                                                                     крестика

Найти:                      Q = x = 306,4·4/3·22,4 = 25,5 кДж

m(O₂)= x гр

                                 Ответ: 25,5 кДж.

Пояснення:

Answer 2

Calculation of Reactant Volume

When pentane is burned, the balanced chemical equation is:

C5H12 + 8O2 → 5CO2 + 6H2O

From the balanced equation, it's clear that 1 mole of pentane reacts with 8 moles of oxygen to produce 5 moles of carbon dioxide and 6 moles of water.

Given that the volume of carbon dioxide produced is 8.96 L, we can use the ideal gas law to find the volume of the reactant. The molar volume of any gas at standard temperature and pressure (STP) is 22.4 L/mol.

Using the relationship between moles and volume at STP, we can calculate the moles of carbon dioxide produced and then use the stoichiometry of the balanced equation to find the moles of pentane reacted.

Let's calculate the moles of carbon dioxide produced first:

moles of CO2 = volume of CO2 / molar volume at STP moles of CO2 = 8.96 L / 22.4 L/mol moles of CO2 ≈ 0.4 moles

Now, using the stoichiometry of the balanced equation, we can find the moles of pentane reacted:

moles of C5H12 = moles of CO2 / 5 moles of C5H12 = 0.4 moles / 5 moles of C5H12 = 0.08 moles

Finally, we can find the volume of pentane reacted using the molar volume at STP:

volume of C5H12 = moles of C5H12 * molar volume at STP volume of C5H12 = 0.08 moles * 22.4 L/mol volume of C5H12 = 1.792 L

Therefore, the volume of the pentane that reacted is approximately 1.792 liters.