What amount of lithium should be taken so that when it interacts with 200 ml of water, a 5% solution

What amount of lithium should be taken so that when it interacts with 200 ml of water, a 5% solution of lithium hydroxide is formed?

200 ml of water – 200 g of water.
Let’s write down the reaction equations:
2Li + 2 H2O = 2LiOH + H2 ↑.
Let 2 * x mol of lithium entered into the reaction, and 2 * x salt of lithium hydroxide and x mol – hydrogen were formed.
Let’s write an expression to determine the amount of a substance:
ν = m / M, whence:
m = v * M.
Let’s define the molar masses:
M ((LiOH) = 7 + 16 + 1 = 24 g / mol.
M ((Li) = 7 g / mol.
M ((H2) = 2 g / mol.
m (Li) = 2x * 7.
m (H2) = x * 2.
Lithium hydroxide mass:
m (LiOH) = 2 * x * 24 = 48x.
Solution weight:
m (solution) = 200 + m (Li) -m (H2) = 200 + 2x * 7-x * 2 = 200 + 14x-2x = 200 + 12x.
Mass fraction of lithium hydroxide:
w (LaOH) = (48x / (200 + 12x)) * 100
(48x / (200 + 12x)) = 0.05
x = 0.21 mol. Lithium entered 2x mol, which means that 0.42 mol was needed for Lithium.
Answer: 0.42 mol of lithium.