Calculate the amount of copper (II) hydroxide substance that can be released when 20 g of sodium hydroxide and 32 g

| May 30, 2021 | education

Calculate the amount of copper (II) hydroxide substance that can be released when 20 g of sodium hydroxide and 32 g of copper (II) sulfate react. Which of the starting materials is taken in excess?

Given:

m (NaOH) = 20 g

m (CuSO4) = 32 g

To find:

1) What substance is given in excess?

2) n (Cu (OH) 2) -?

Decision:

2NaOH + CuSO4 = Na2SO4 + Cu (OH) 2, – we solve the problem, relying on the composed reaction equation:

1) Find the amount of sodium hydroxide and copper sulfate:

n (NaOH) = m: M = 20 g: 40 g / mol = 0.5 mol

n (CuSO4) = m: M = 32 g: 160 g / mol = 0.2 mol

Since there was more sodium hydroxide, it means that this substance was given in excess. We work with the substance given in the deficiency in order to get more accurate calculations:

2) We compose a logical expression:

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

then 0.2 mol CuSO4 will give x mol Cu (OH) 2,

then x = 0.2 mol.

Answer: NaOH was given in excess; n (Cu (OH) 2) = 0.2 mol.



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