We are Given:
22.5 grams of NaOH in 75 grams of H₂O
2NaOH + 2H₂O → 2Na(OH)₂ + H₂
from this equation, we can see that for 2 moles of both NaOH and H₂O, we get 2 Moles of Na(OH)₂
Calculating the number of moles of NaOH and H₂O provided:
Moles of NaOH:
molar mass of NaOH = 40 grams/mol
Given mass of NaOH = 22.5 grams
Number of moles of NaOH = Given mass / Molar mass
Number of moles of NaOH = 22.5 / 40
Number of moles of NaOH = 1.125 moles of NaOH
Moles of H₂O:
molar mass of H₂O = 18 grams / mol
given mass of H₂O = 75 grams
Number of moles of H₂O = Given mass / Molar mass
Number of moles of H₂O = 75 / 18 = 4.167 moles of H₂O
Number of Moles of NaOH and H₂O which will actually react:
from the equation, we know that the moles of NaOH and H₂O react in 1:1 ratio
So, we can say that one mole of NaOH reacts with one mole of H₂O
but we have much less NaOH than H₂O
So, only the moles equal to the number of moles of NaOH will react of either compound
1.125 moles of both NaOH and H₂O will react
Moles of Na(OH)₂ formed:
we know from the equation that the moles of NaOH and H₂O react to form moles of Na(OH)₂ in the ratio 1:1:1
So, 1.125 moles of Na(OH)₂ will be formed
Percentage composition by mass:
amount of water present alongside the Na(OH)₂ formed:
we were given 4.167 moles of H₂O and only 1.125 moles of it was consumed in the reaction
so, 4.167 - 1.125 = 3 moles of H₂O will be present alongside the Na(OH)₂
3 moles of H₂O = 3 * 18 = 54 grams of H₂O will be present along the Na(OH)₂
Mass of Na(OH)₂ formed:
mass = number of moles * molar mass
mass = 1.125 * 41 = 46 grams of Na(OH)₂ will be formed
Total mass of the solution formed:
total mass = mass of H₂O + mass of Na(OH)₂
total mass = 54 + 46 = 100 grams
Percent composition by mass of Na(OH)₂:
Mass% of Na(OH)₂ = (Mass of Na(OH)₂ / Total mass) * 100
Mass% of Na(OH)₂ = 46 %