Electrochemistry metallurgy mass calculation

[ambition218]

Homework Statement

Calculate the minimum mass of carbon needed to reduce 1.00 kg from sufficient zinc oxide.

Homework Equations

The Attempt at a Solution

I think the equation would be 2 ZnO + C = Zn + CO2
I know the molar mass of C is 12.01, O is 16 and Zn is 65.4

I have been taking Chemisty by distance and nowhere in my text materials can I find how to solve this question.

 

[Borek]

It is a simple stoichiometry. First, balance your reaction equation - you already have correct reactants and products. Do you know how to read reaction equation?

 

[shikobe]

Good guess on the chemical equation...but this type of reaction is smelting of a metal oxide via a heat furnace, which is done industrially. The products for this reaction is not CO2 but is CO, carbon monoxide. The reaction should look like: ZnO+C>Zn+CO

You need to take mole ratios of what is given...a trick is sometimes if you forget a formula is to use the units given and then play with them to get what you want.

 

[Borek]

From what I understand CO is still strong enough reducing agent to reduce ZnO, so both products (CO & CO₂) are possible.

 

[DeeNos]

I can help you with this problem. First, we need to determine the amount of zinc oxide (ZnO) that needs to be reduced. Since we have 1.00 kg of ZnO, we can convert this into moles using its molar mass, which is 81.4 g/mol. This gives us 1000 g / 81.4 g/mol = 12.28 moles of ZnO.

Next, we need to determine the amount of carbon (C) needed to reduce this amount of ZnO. From the balanced chemical equation you provided, we know that 2 moles of ZnO requires 1 mole of C. Therefore, we need half the amount of moles of C compared to ZnO. This means we need 12.28 moles / 2 = 6.14 moles of C.

Finally, we can calculate the mass of carbon needed by multiplying the moles of C by its molar mass, which is 12.01 g/mol. This gives us 6.14 moles * 12.01 g/mol = 73.6 g of C.

Therefore, the minimum mass of carbon needed to reduce 1.00 kg of ZnO is 73.6 g. I hope this helps with your homework problem!