Calculate the mass fraction of the yield of copper hydroxide, when reacting 22.4 g of potassium hydroxide

Calculate the mass fraction of the yield of copper hydroxide, when reacting 22.4 g of potassium hydroxide with copper sulfate, if 15 g of copper hydroxide is obtained during the reaction.

Let’s implement the solution:
1. Let’s write down the equation and arrange the coefficients:
2KOH + CuSO4 = Cu (OH) 2 + K2SO4 – ion exchange reaction, copper hydroxide precipitate was formed;
2. Find the molar masses of substances:
M (KOH) = 39.1 + 16 + 1 = 56.1 g / mol;
M Cu (OH) 2 = 63.5 + (16 + 1) * 2 = 97.5 g / mol;
3. Let’s calculate the number of mol of potassium hydroxide, if the mass is known:
Y (KOH) = m / M = 22.4 / 56.1 = 0.339 mol;
4. Let’s make the proportion using the given equations:
0.339 mol (KOH) – X mol Cu (OH) 2;
2 mol -1 mol hence, X mol Cu (OH) 2 = 0.339 * 1/2 = 0.169 mol;
5. Determine the theoretical mass of copper hydroxide:
m (theoretical) = Y * M = 0.169 * 97.5 = 16.47 g;
6. Let’s calculate the mass fraction of the output of copper hydroxide by the formula:
W = m (practical) / m (theoretical) * 100 = 15 / 16.47 = 91.07%.
Answer: The mass fraction of the output of copper hydroxide is 91.07%.