Question

. In the production of copper from ore containing copper(II) sulfide, the ore is first roasted to change it to the oxide according to the following equation: 2CuS 3O2 £ 2CuO 2SO2 a. If 100 g of CuS and 56 g of O2 are available, which reactant is limiting? b. What mass of CuO can be formed from the reaction of 18.7 g of CuS and 12.0 g of O2? 4. A reaction such as

Answer 1

(a) The limiting reactant is, CuS

(b) The mass of
`CuO` produced is, 15.6 grams.

Explanation :

Part (a) :

Given,

Mass of
`CuS` = 100 g

Mass of
`O_2` = 56 g

Molar mass of
`CuS` = 95.6 g/mol

Molar mass of
`O_2` = 32 g/mol

First we have to calculate the moles of
`CuS` and
`O_2`.

`\text{Moles of }CuS=\frac{\text{Given mass }CuS}{\text{Molar mass }CuS}`

`\text{Moles of }CuS=(100g)/(95.6g/mol)=1.05mol`

and,

`\text{Moles of }O_2=\frac{\text{Given mass }O_2}{\text{Molar mass }O_2}`

`\text{Moles of }O_2=(56g)/(32g/mol)=1.75mol`

Now we have to calculate the limiting and excess reagent.

The balanced chemical equation is:

`2CuS+3O_2\rightarrow 2CuO+2SO_2`

From the balanced reaction we conclude that

As, 2 mole of
`CuS` react with 3 mole of
`O_2`

So, 1.05 moles of
`CuS` react with
`(3)/(2)* 1.05=1.58` moles of
`O_2`

From this we conclude that,
`O_2` is an excess reagent because the given moles are greater than the required moles and
`CuS` is a limiting reagent and it limits the formation of product.

Hence, the limiting reactant is, CuS

Part (b) :

Given,

Mass of
`CuS` = 18.7 g

Mass of
`O_2` = 12.0 g

Molar mass of
`CuS` = 95.6 g/mol

Molar mass of
`O_2` = 32 g/mol

First we have to calculate the moles of
`CuS` and
`O_2`.

`\text{Moles of }CuS=\frac{\text{Given mass }CuS}{\text{Molar mass }CuS}`

`\text{Moles of }CuS=(18.7g)/(95.6g/mol)=0.196mol`

and,

`\text{Moles of }O_2=\frac{\text{Given mass }O_2}{\text{Molar mass }O_2}`

`\text{Moles of }O_2=(12.0g)/(32g/mol)=0.375mol`

Now we have to calculate the limiting and excess reagent.

The balanced chemical equation is:

`2CuS+3O_2\rightarrow 2CuO+2SO_2`

From the balanced reaction we conclude that

As, 2 mole of
`CuS` react with 3 mole of
`O_2`

So, 0.196 moles of
`CuS` react with
`(3)/(2)* 0.196=0.294` moles of
`O_2`

From this we conclude that,
`O_2` is an excess reagent because the given moles are greater than the required moles and
`CuS` is a limiting reagent and it limits the formation of product.

Now we have to calculate the moles of
`CuO`

From the reaction, we conclude that

As, 2 mole of
`CuS` react to give 2 mole of
`CuO`

So, 0.196 mole of
`CuS` react to give 0.196 mole of
`CuO`

Now we have to calculate the mass of
`CuO`

`\text{ Mass of }CuO=\text{ Moles of }CuO* \text{ Molar mass of }CuO`

Molar mass of
`CuO` = 79.5 g/mole

`\text{ Mass of }CuO=(0.196moles)* (79.5g/mole)=15.6g`

Therefore, the mass of
`CuO` produced is, 15.6 grams.