Question

Химия: Дано: m ( Li )- 15 г H2O Найти: V ( H2)-? m (LiOH)-?

Answer 1

2,14 моль
2Li + 2H2O = 2LiOH + H2
2 моль           2 моль   1 моль

M(LiOH) = 7 + 16 + 1 = 24 г/моль
n = m/M = V/Vм, Vм = 22,4 л/моль
(при н.у., любого газа)

n(Li) = 15/7 = 2,14 моль = n(LiOH)
m(LiOH) = 2,14*24 = 51,36 г
по уравнению реакции:
n(H2) = 2,14/2 = 1,07 моль
V(H2) = 1,07*22,4 = 23,97 л

Answer 2

Given:

- m(Li) = 15 g - H2O

To find:

- V(H2) - m(LiOH)

Solution:

To find the volume of hydrogen gas (V(H2)) produced, we need to use the stoichiometry of the reaction between lithium (Li) and water (H2O). The balanced chemical equation for this reaction is:

2Li + 2H2O -> 2LiOH + H2

From the balanced equation, we can see that 2 moles of Li react with 2 moles of H2O to produce 1 mole of H2. This means that the mole ratio between Li and H2 is 2:1.

To find the moles of Li, we can use the given mass and the molar mass of Li. The molar mass of Li is approximately 6.94 g/mol. Therefore, the number of moles of Li (n(Li)) can be calculated as follows:

n(Li) = m(Li) / M(Li) = 15 g / 6.94 g/mol ≈ 2.16 mol

Since the mole ratio between Li and H2 is 2:1, the number of moles of H2 (n(H2)) produced will be half of the number of moles of Li:

n(H2) = 1/2 * n(Li) = 1/2 * 2.16 mol ≈ 1.08 mol

To find the volume of H2 gas (V(H2)), we can use the ideal gas law equation:

PV = nRT

Assuming the reaction takes place at standard temperature and pressure (STP), we can use the values: - P = 1 atm - T = 273 K - R = 0.0821 L·atm/(mol·K)

Substituting the values into the equation, we can solve for V(H2):

V(H2) = n(H2) * RT / P = 1.08 mol * 0.0821 L·atm/(mol·K) * 273 K / 1 atm ≈ 24.2 L

Therefore, the volume of H2 gas produced (V(H2)) is approximately 24.2 L.

To find the mass of lithium hydroxide (m(LiOH)) produced, we need to use the stoichiometry of the reaction. From the balanced equation, we can see that 2 moles of Li react to produce 2 moles of LiOH. This means that the mole ratio between Li and LiOH is 2:2, or simply 1:1.

Since the number of moles of Li (n(Li)) is approximately 2.16 mol, the number of moles of LiOH (n(LiOH)) produced will also be approximately 2.16 mol.

To find the mass of LiOH, we can use the molar mass of LiOH, which is approximately 23.95 g/mol. Therefore, the mass of LiOH (m(LiOH)) can be calculated as follows:

m(LiOH) = n(LiOH) * M(LiOH) = 2.16 mol * 23.95 g/mol ≈ 51.6 g

Therefore, the mass of lithium hydroxide produced (m(LiOH)) is approximately 51.6 g.