To solve, we write down the equation, select the coefficients:
K2O + H2O = 2KOH – hydration reaction, obtained potassium hydroxide;
M (K2O) = 94.2 g / mol;
M (KOH) = 56.1 g / mol.
Determine the amount of mol of potassium oxide, if the mass is known:
Y (K2O) = m / M = 47 / 94.2 = 0.498 mol.
Let’s make the proportion:
0.498 mol (K2O) – X mol (KOH);
-1 mol -2 mol hence, X mol (KOH) = 0.498 * 2/1 = 0.997 mol.
Find the theoretical mass of potassium hydroxide:
m (KOH) = Y * M = 0.997 * 56.1 = 55.97 g.
Let’s calculate the mass fraction of the product yield:
W = m (practical) / m (theoretical) * 100 = 50 / 55.97 * 100 = 89.33%.
Answer: W (KOH) is 89.33%.
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