Question

At very high temperatures, copper(II) sulfate undergoes the reaction CuSO4(s) --> CuO (s) + SO3 (g)

Answer 1

Answer : This is a type of decomposition reaction. The mass of copper oxide produced would be, 159 grams

Solution : Given,

Mass of copper sulfate = 319.22 g

Molar mass of copper sulfate = 159.6 g/mole

Molar mass of copper oxide = 79.5 g/mole

First we have to calculate the moles of copper sulfate.

$\text{Moles of }CuSO_4=\frac{\text{Mass of }CuSO_4}{\text{Molar mass of }CuSO_4}=\frac{319.22g}{159.6g/mole}=2moles$

Now we have to calculate the moles of copper oxide.

The balanced chemical reaction is,

$CuSO_4(s)\rightarrow CuO(s)+SO_3(g)$

This reaction is a decomposition reaction in which the larger reactant molecule decomposes to give two or more new products.

From the reaction, we conclude that

As, 1 mole of copper sulfate decomposes to produce 1 mole of copper oxide

So, 2 moles of copper sulfate decomposes to produce 2 moles of copper oxide

Now we have to calculate the mass of copper oxide.

$\text{Mass of }CuO=\text{Moles of }CuO\times \text{Molar mass of }CuO$

$\text{Mass of }CuO=(2mole)\times (79.5g/mole)=159g$

Therefore, the mass of copper oxide produced would be, 159 grams

Answer 2

Answer: a) Decomposition

b) 159.0 g of CuO(s) would be produced.

Explanation:

Decomposition reaction is a chemical reaction in which one reactant gives two or more than two products. They are endothermic reactions as energy is required to break the bonds of the reactants.

$CuSO_4(s)\rightarrow CuO(s)+SO_3(g)$

As can be seen from the given balanced reaction:  

1 mole of $CuSO_4$ gives 1 mole of $CuO$.

Thus 159.6 g of  $CuSO_4$ gives 79.5 g of $CuO$.

319.22 g of  $CuSO_4$ will give=$\frac{79.5}{159.6}\times 319.22=159.0g$of $CuO$.