Question

If 2.50 metric tons of bornite is reacted with excess O2 and the process has an 86.3% yield of copper, what mass of copper is produced?

Answer 1

Answer : The mass of copper produced is,
`1.20\text{ metric ton}`

Explanation : Given,

Mass of bornite
`Cu_3FeS_3` =
`2.50\text{ metric ton}=2.50* 10^6g`

conversion used :
`1\text{ metric ton}=10^6g`

Molar mass of
`Cu_3FeS_3` = 342.678 g/mole

First we have to calculate the moles of
`Cu_3FeS_3`.

`\text{Moles of }Cu_3FeS_3=\frac{\text{Mass of }Cu_3FeS_3}{\text{Molar mass of }Cu_3FeS_3}=(2.50* 10^6g)/(342.678g/mole)=7295.48mole`

Now we have to calculate the moles of
`Cu`.

The balanced chemical reaction will be,

`2Cu_3FeS_3+7O_2\rightarrow 6Cu+2FeO+6SO_2`

From the balanced reaction, we conclude that

As, 2 moles of
`Cu_3FeS_3` react to give 6 moles of
`Cu`

So, 7295.48 moles of
`Cu_3FeS_3` react to give
`(6)/(2)* 7295.48=21886.44` moles of
`Cu`

Now we have to calculate the mass of
`Cu`

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

`\text{Mass of }Cu=(21886.44mole)* (63.546g/mole)=1390795.716g`

The theoretical yield of
`Cu` = 1390795.716 g

The actual yield of
`Cu` = ?

Now we have to calculate the actual yield of
`Cu`.

`\%\text{ yield of }Cu=\frac{\text{Actual yield of }Cu}{\text{Theoretical yield of }Cu}* 100`

`86.3=\frac{\text{Actual yield of }Cu}{1390795.716g}* 100`

`\text{Actual yield of }Cu=1200256.703g=1.20* 10^6g=1.20\text{ metric ton}`

Therefore, the mass of copper produced is,
`1.20\text{ metric ton}`