25.07grams is the theoretical yield of magnesium hydroxide if the resulting solution has a volume of 2. 82 liters.
<h3>How we calculate the mass from moles?</h3>
Mass of any substance will be calculated from moles as:
n = W/M, where
W = given or required mass
M = molar mass
Given chemical reaction is:
2NaOH + MgCl₂ → Mg(OH)₂ + 2NaCl
From the stoichiometry of the reaction, it is clear that:
2 moles of NaOH = produce 1 mole of Mg(OH)₂
1 mole of NaOH = produce 1/2 mole of Mg(OH)₂
Given concentration of NaOH = 1.35M
Given volume of NaOH = 637mL = 0.637L
Moles of NaOH will be calculated as:
M = n/V
n = 1.35M × 0.637L = 0.859 moles
So, 0.859 moles of NaOH = produce 0.859×1/2 = 0.429 moles of Mg(OH)₂
Now, we convert this moles into grams by using the above formula:
0.429 = W/58.31 g/mole
W = 25.07grams
Hence, 25.07grams is the theoretical yield of magnesium hydroxide.
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