Question

Copper sulphate solution reacts with sodium hydroxide solution to form a precipitate of copper hydroxide according to the equation:
$2NaOH+CuS{{O}_{4}}\to N{{a}_{2}}S{{O}_{4}}+Cu{{(OH)}_{2}}\downarrow $
(i)what mass of copper hydroxide is precipitated by using 200 gm of sodium hydroxide?
[H = 1, O = 16, Na = 23, S = 32, Cu = 64]

Answer 1

Hint: precipitation is often a result in the double displacement reactions. The number of moles in a balanced chemical equation gives us information about the mass of the species involved. 1 mole of any compound carries a mass equal to the gram molecular mass of that compound.

Complete answer:
We have been given a reaction, where copper sulphate solution is mixed with sodium hydroxide, which forms sodium sulphate and a precipitate of copper hydroxide. we have to find the mass of copper hydroxide which is precipitated in this reaction.
From the chemical equation and the given atomic masses, we will take out the masses of all the compounds using stoichiometry of the given equation, and then by unitary method we will calculate the mass of copper hydroxide for 200 g of sodium hydroxide.
As, given in the equation, 2 moles of sodium hydroxide gives 1 mole of copper sulphate. This means 80 g of sodium hydroxide, is giving 98 g of copper hydroxide. So, we will calculate the mass in 200 g of sodium hydroxide as,
Mass of $Cu{{(OH)}_{2}}$in 200 g NaOH = $\dfrac{98\,g\,Cu{{(OH)}_{2}}}{80\,g\,NaOH}\times 200\,g\,NaOH$
Mass of $Cu{{(OH)}_{2}}$in 200 g NaOH = 245 g
Hence, the mass of $Cu{{(OH)}_{2}}$is 245 g in 200 g NaOH.

Note:
The molar mass is calculated by the given masses, as Na + O + H = 23 + 16 + 1 = 40 g. While for $Cu{{(OH)}_{2}}$= Cu + 2(O+H) = 64 + 2(16+1) = 98 g. These are the masses given in 1 gram of any atom, so they are gram atomic masses. While, 1 mole of any atom contains mass equal to its atomic mass and same for molecules.