Question

Помогите с решением задачи,желательно все расписать и полное решением Задача. Определите обьем кислорода,который вступит в реакцию с литием,если в результате реакции получиться оксид лития массой 12 грамм. За подробное решение +100 к карме

Answer 1

Дано:
m(Li₂O)= 12 грамм
Найти:
V(O₂) - ?
Решение:
4Li + O₂ → 2Li₂O
Mr(Li₂O) = 7*2+16=30 г\моль
m=M*n ⇒ n= m\M
n(Li₂O) = 12 г : 30г\моль = 0.4 моль
n(O₂) = n(Li₂O)
  1      :      2
n(O₂) = 0,4 \ 2 = 0,2 моль
V= Vm*n
V= 22,4 л\моль * 0,2 моль = 4,48 л
Ответ: V(O₂) = 4, 48 л

Answer 2

Problem:

We need to determine the volume of oxygen that will react with lithium to produce lithium oxide with a mass of 12 grams.

Solution:

To solve this problem, we can use the stoichiometry of the reaction between lithium and oxygen. The balanced chemical equation for the reaction is:

2Li + O2 -> 2Li2O

From the balanced equation, we can see that 2 moles of lithium react with 1 mole of oxygen to produce 2 moles of lithium oxide.

To find the volume of oxygen, we need to convert the mass of lithium oxide to moles using its molar mass, and then use the mole ratio from the balanced equation to find the moles of oxygen. Finally, we can use the ideal gas law to convert the moles of oxygen to volume.

Let's go step by step:

Step 1: Calculate the moles of lithium oxide. The molar mass of lithium oxide (Li2O) is the sum of the molar masses of lithium (Li) and oxygen (O). The molar mass of lithium is approximately 6.94 g/mol, and the molar mass of oxygen is approximately 16.00 g/mol.

Molar mass of Li2O = (2 * molar mass of Li) + molar mass of O = (2 * 6.94 g/mol) + 16.00 g/mol = 13.88 g/mol + 16.00 g/mol = 29.88 g/mol

Now, we can calculate the moles of lithium oxide using its mass:

Moles of Li2O = Mass of Li2O / Molar mass of Li2O = 12 g / 29.88 g/mol ≈ 0.401 moles

Step 2: Calculate the moles of oxygen. From the balanced equation, we know that 2 moles of lithium react with 1 mole of oxygen to produce 2 moles of lithium oxide.

Moles of O2 = (Moles of Li2O / 2) * (1 mole of O2 / 2 moles of Li2O) = (0.401 moles / 2) * (1 mole / 2 moles) = 0.2005 moles

Step 3: Calculate the volume of oxygen. To calculate the volume of oxygen, we can use the ideal gas law, which states:

PV = nRT

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

Rearranging the equation to solve for V, we have:

V = (nRT) / P

Substituting the values:

V = (0.2005 moles * 0.0821 L·atm/(mol·K) * 298 K) / 1 atm ≈ 4.92 L

Therefore, the volume of oxygen that will react with lithium to produce lithium oxide with a mass of 12 grams is approximately 4.92 liters.

Please note that this calculation assumes ideal gas behavior and standard conditions.