Calculate the mass of the precipitate that precipitated from the interaction of 980 g

| July 22, 2021 | education

Calculate the mass of the precipitate that precipitated from the interaction of 980 g of a 20% solution of copper sulfate (2) with potassium hydroxide.

Given:

m (CuSO4) = 980 g

w% (CuSO4) = 20%

To find:

m (draft) -?

Decision:

CuSO4 + 2KOH = K2SO4 + Cu (OH) 2, – we solve the problem, relying on the composed reaction equation:

1) Find the mass of copper sulfate in the reacted solution:

m (CuSO4) = 980 g * 0.2 = 196 g

2) Find the amount of copper sulfate:

n (CuSO4) = m: M = 196 g * 160 g / mol = 1.225 mol

3) We compose a logical expression:

If 1 mol of CuSO4 gives 1 mol of Cu (OH) 2,

then 1.225 mol of CuSO4 will give x mol of Cu (OH) 2,

then x = 1.225 mol.

4) Find the mass of copper hydroxide precipitated during the reaction:

m (Cu (OH) 2) = n * M = 1.225 mol * 98 g / mol = 120.05 g.

Answer: m (Cu (OH) 2) = 120.05 g.



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