How many grams of precipitate is formed when 160 grams of CuSO4 interacts with the required amount of NaOH ?

| August 16, 2021 | education

Let’s implement the solution:
1. Write down the equation according to the problem statement:
m = 160 g. X g. -?
CuSO4 + 2NaOH = Cu (OH) 2 + Na2SO4 – ion exchange reaction, a precipitate of copper hydroxide, sodium sulfate was obtained;
2. Let’s calculate the molar masses of inorganic substances:
M (CuSO4) = 159.5 g / mol;
M Cu (OH) 2 = 97.5 g / mol.
3. Calculate the number of moles of the starting material, if the mass is known:
Y (CuSO4) = 160 / 159.5 = 1 mol;
Y Cu (OH) 2 = 1 mol since the amount of these substances according to the equation is equal to 1 mol.
4. Find the mass of the reaction product:
m Cu (OH) 2 = Y * M = 1 * 97.5 = 97.5 g.
Answer: in the course of the ion exchange reaction, copper hydroxide with a mass of 97.5 g was obtained.



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