Câu 25: Trong 1 bình kín dung tích không đới chứa 5,12 gam khí SO_(2) và 1,28 gam khiO_(2). Thực hiện phản ứng tổng SO_(3)(V_(2)O_(5)) Sau đó đưa bình về nhiệt độ ban đầu. Khi phản ứng đạt tới trạng thái cân bằng, lượng khí SO_(2) còn lại bằng 25% so với lượng ban đầu.Nếu áp suất ban đầu là 3 atm thì áp suất lúc cân bằng là A. 2,3 atm. B. 2,2 atm. C. 2,1 atm. D. 2,5 atm.

<p>This question is a typical problem of equilibrium in chemical reactions. The reaction is SO2 + 1/2 O2 -> SO3, which occurs in a closed container. Initially, the container has 5.12 grams of SO2 and 1.28 grams of O2. After the reaction, 25% of the initial SO2 remains, which means that 75% of it reacts. The pressure of the gases also changes during the reaction.<br /><br />Initially, from the given masses and molar masses of the gases, the number of moles of SO2 and O2 are calculated. From the reaction stoichiometry, the change in the number of moles of each gas is calculated when 75% of SO2 reacts.<br /><br />As the reaction takes place in a closed container and the temperature is the same at the start and the equilibrium, the pressure is proportional to the number of moles of the gases (from the ideal gas law). Hence, the initial pressure is multiplied by the ratio of the total moles of gases at equilibrium to the total moles of gases initially to get the pressure at equilibrium.<br /><br />Through the above analysis and calculations, we get the equilibrium pressure to be 2.2 atm. So, the answer is B.</p>