Hấp thụ hoàn toàn 2,688 lít CO2 (đktc) vào V lít dung dịch Ba(OH)2 nồng độ 0,4 M thu được 15,76 gam kết tủa trắng. Tính V.
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Phương pháp giải:

Bước 1: Viết phương trình phản ứng:
CO2 + 2Ba(OH)2 → BaCO3 + 2H2O

Bước 2: Tính số mol CO2 đã hấp thụ:
Số mol (CO2) = khối lượng (CO2) / khối lượng riêng (CO2)
Số mol (CO2) = 2,688 lít / 22,4 lít (đktc)

Bước 3: Từ phương trình phản ứng, ta biết rằng 1 mol CO2 tương ứng với 1 mol BaCO3. Do đó, số mol BaCO3 = số mol CO2.

Bước 4: Tính khối lượng BaCO3 tạo thành:
khối lượng (BaCO3) = số mol (BaCO3) x khối lượng riêng (BaCO3)
khối lượng (BaCO3) = số mol (CO2) x khối lượng riêng (BaCO3)

Bước 5: Tìm dung tích V:
Công thức tính nồng độ = số mol chất / dung tích chất (trong lít)
0,4 M = số mol Ba(OH)2 / V
V = số mol Ba(OH)2 / 0,4 M

Bước 6: Tìm số mol Ba(OH)2:
số mol (Ba(OH)2) = số mol (BaCO3) / 2

Bước 7: Tính dung tích V:
V = (số mol (BaCO3) / 2) / 0,4 M

Bước 8: Tính giá trị V:
V = [số mol (CO2) x khối lượng riêng (BaCO3)] / [2 x 0,4 M]

Câu trả lời:
Giá trị của V là kết quả tính toán từ phương pháp giải trên. V tương ứng với dung tích dung dịch Ba(OH)2 đã hấp thụ hoàn toàn CO2 để thu được 15,76 gam kết tủa trắng.

According to the balanced chemical equation Ba(OH)2 + CO2 -> BaCO3 + H2O, we can determine the molar ratio between Ba(OH)2 and BaCO3 is 1:1. The molar mass of BaCO3 is 197.34 g/mol. From the given information, we know that the mass of BaCO3 formed is 15.76 grams. Using the molar mass, we can calculate the number of moles of BaCO3 as follows: 15.76 g BaCO3 * (1 mole BaCO3 / 197.34 g BaCO3) = 0.0798 moles of BaCO3. As the molar ratio is 1:1, the number of moles of Ba(OH)2 used is also 0.0798 moles. Since the molar concentration of Ba(OH)2 solution is 0.4 M, we can calculate the volume of Ba(OH)2 solution used as follows: V = n / C = 0.0798 moles / 0.4 mol/L = 0.1995 L = 1.995 * 10^-1 L.

To find the volume of the Ba(OH)2 solution, we need to determine the number of moles of CO2 using the given information. The molar mass of CO2 is 44.01 g/mol. From the given information, the amount of BaCO3 formed is 15.76 grams. Using the molar mass, we can calculate the number of moles of BaCO3 as follows: 15.76 g BaCO3 * (1 mole BaCO3 / 197.34 g BaCO3) = 0.0798 moles of BaCO3. As 1 mole of Ba(OH)2 reacts with 1 mole of CO2, the number of moles of CO2 is also 0.0798 moles. Using the ideal gas law equation (PV = nRT), where P is the pressure, V is the volume, n is the number of moles, R is the ideal gas constant, and T is the temperature, we can calculate the volume of CO2 at standard conditions as follows: V = nRT/P = 0.0798 moles * 0.0821 L*atm/(mol*K) * 273.15 K / 1 atm = 1.760 L. Therefore, the volume of the Ba(OH)2 solution is 1.760 liters.

The molar mass of Ba(OH)2 is 171.34 g/mol. From the given information, we know that the molar ratio between Ba(OH)2 and BaCO3 is 1:1. Therefore, the number of moles of BaCO3 formed is equal to the number of moles of Ba(OH)2 used. Since the molar concentration of Ba(OH)2 solution is 0.4 M, the number of moles of Ba(OH)2 used can be calculated as follows: x moles Ba(OH)2 = 0.4 M * V liters. Since the molar ratio is 1:1, the number of moles of BaCO3 formed is also x moles. From the given information, we can calculate the mass of BaCO3 formed as follows: x moles * molar mass of BaCO3 = 15.76 g. Rearranging the equation, we have x = 15.76 g / molar mass of BaCO3. Substituting the value of molar mass of BaCO3, we can calculate x = 0.0798 moles. Since the number of moles of Ba(OH)2 used is equal to the number of moles of CO2 present, we can calculate the volume of the Ba(OH)2 solution as follows: V = x moles / molar concentration = (0.0798 moles) / (0.4 mol/L) = 0.1995 L = 1.995 * 10^-1 L.

The balanced chemical equation for the reaction between Ba(OH)2 and CO2 is Ba(OH)2 + CO2 -> BaCO3 + H2O. According to the equation, 1 mole of Ba(OH)2 reacts with 1 mole of CO2 to form 1 mole of BaCO3. From the given information, we can calculate the number of moles of BaCO3 formed as follows: 15.76g BaCO3 * (1 mole BaCO3 / molar mass of BaCO3) = x moles BaCO3. Since the molar mass of BaCO3 is 197.34 g/mol, x = 0.0798 moles BaCO3. Since 1 mole of Ba(OH)2 reacts with 1 mole of CO2, the number of moles of CO2 present is also 0.0798 moles. As the volume of CO2 is given as 2.688 liters at standard conditions, we can use the ideal gas law equation to determine the molar volume of CO2: V = n * (RT/P), where R is the ideal gas constant, T is the temperature in Kelvin, and P is the pressure in atmospheres. Assuming standard conditions (T = 273.15 K and P = 1 atm), we can calculate V as follows: V = 0.0798 moles * (0.0821 L*atm/(mol*K)) * 273.15 K / 1 atm = 1.760 L. Therefore, the volume of the Ba(OH)2 solution is 1.760 liters.