Question

Iron reacts with copper (I) sulfate to produce copper and iron (II) sulfate. If 4.32x10^25

particles of copper (I) sulfate react, how many moles of copper will be produced?
Fe + Cu2504 --> FeSO4 + 2Cu

Answer 1

Answer: 143.48 moles of copper will be produced.

Step-by-step explanation:

We are given:

Number of particles of copper (I) sulfate =
`4.32* 10^(25)`

According to the mole concept:

`6.022* 10^(23)` number of particles is contained in 1 mole of a compound

So,
`4.32* 10^(25)` number of particles will be contained in =
`(1)/(6.022* 10^(23))* 4.32* 10^(25)=71.74mol` of copper (I) sulfate

The given chemical equation follows:

`Fe+Cu_2SO_4\rightarrow FeSO_4+2Cu`

By the stoichiometry of the reaction:

If 1 mole of copper (I) sulfate produces 2 moles of Cu

So, 71.74 moles of copper (I) sulfate will produce =
`(2)/(1)* 71.74=143.48mol` of Cu

Hence, 143.48 moles of copper will be produced.