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Please could you help me with this calculation :

A blast furnace can produce about 700 tonnes of iron a day. How much iron (III) oxide will be consumed? Assuming coke is pure carbon, how much coke would be needed to produce the necessary carbon monoxide?

part a) Fe2O3(s) + 3CO(g) = 2Fe(l) + 3CO2(g)

My answer was 350 tonnes, I used ratio

I don't know how to do part b: 2C(s) + O2(g)= 2CO(g)

Thanks

A blast furnace can produce about 700 tonnes of iron a day. How much iron (III) oxide will be consumed? Assuming coke is pure carbon, how much coke would be needed to produce the necessary carbon monoxide?

part a) Fe2O3(s) + 3CO(g) = 2Fe(l) + 3CO2(g)

My answer was 350 tonnes, I used ratio

I don't know how to do part b: 2C(s) + O2(g)= 2CO(g)

Thanks

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#2

(Original post by

Please could you help me with this calculation :

A blast furnace can produce about 700 tonnes of iron a day. How much iron (III) oxide will be consumed? Assuming coke is pure carbon, how much coke would be needed to produce the necessary carbon monoxide?

part a) Fe2O3(s) + 3CO(g) = 2Fe(l) + 3CO2(g)

My answer was 350 tonnes, I used ratio

I don't know how to do part b: 2C(s) + O2(g)= 2CO(g)

Thanks

**sheepshap**)Please could you help me with this calculation :

A blast furnace can produce about 700 tonnes of iron a day. How much iron (III) oxide will be consumed? Assuming coke is pure carbon, how much coke would be needed to produce the necessary carbon monoxide?

part a) Fe2O3(s) + 3CO(g) = 2Fe(l) + 3CO2(g)

My answer was 350 tonnes, I used ratio

I don't know how to do part b: 2C(s) + O2(g)= 2CO(g)

Thanks

Actually, your answer is not correct.

In such calculations, you must always remember to

**convert the masses into moles first.**

Here, molecular mass of Fe =56 . Mass of iron produced in a day = 700 ton. =? g

So, moles of iron produced in a day = x.

**Calculate x**.[Be sure to convert mass from ton to g]

From the chemical equation, one mole of Fe is seen to be produced from 1/2 mole of Fe2O3.

Hence, x mole Fe will require x/2 mole of Fe2O3.

Finally, mass of Fe2O3 consumed per day = moles x molecular mass

= (x/2) x molecular mass of Fe2O3.

To get the answer, you'd have to calculate the molecular mass of Fe2O3 first.

**You'll get the answer in g.**

The answer is 1000 tons.

**Part (b)**

2C + O2 ---> 2 CO

First, from part (a), we calculate the no. of moles of CO used to obtain 700 tons of iron.

From the first chem equation, 2 moles of iron need 3 moles of CO.

So, one mole of iron needs 3/2 moles of CO.

Therefore, x moles of Fe will be obtained from 3x/2 moles of CO.

From the

**second chemical equation**, we find that one mole of CO is generated from one mole of carbon (coke).

Hence, (3x/2) moles of CO would require (3x/2 ) moles of coke.

But we need the answer in terms of mass, i.e. g.

Mass of coke required = no. of moles x molecular mass

= (3x/2) x molecular mass of coke

= (3x/2) x molecular mass of carbon. Just plug in the value x calculated in part (a) to evaluate this quantity.

**Your answer will be in g.**

Give it a try here. In case you get stuck, somebody will surely help you out.

Good luck !

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#3

**sheepshap**)

Please could you help me with this calculation :

A blast furnace can produce about 700 tonnes of iron a day. How much iron (III) oxide will be consumed? Assuming coke is pure carbon, how much coke would be needed to produce the necessary carbon monoxide?

part a) Fe2O3(s) + 3CO(g) = 2Fe(l) + 3CO2(g)

My answer was 350 tonnes, I used ratio

I don't know how to do part b: 2C(s) + O2(g)= 2CO(g)

Thanks

**A general approach for solving such problems:**

To solve such problems, one needs to go through the following steps:

**Step 1)**Calculate all the molecular masses.

**Step 2)**Convert all the given masses into moles, by dividing each mass by corresponding molecular mass.

**Step 3)**From the chemical equation, look up the molar ratios.

As an example, consider the hypothetical chemical equation:

aA + bB -------> cC + dD

For every d moles of product D generated, b moles of reactant B would be consumed. So, the molar ratio is d:b for this pair. Hence, for every mole of D

formed, b/d mole of B will be used up.

**Step 4)**Convert any mole into mass, by multiplying with the corresponding molecular mass.

**Remember that there is no need to change the unit of mass. Your answer will be in the same unit of mass as in the question. In the problem above, the answer will thus be in tons.**

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#4

how do calculate the mass of iron that could be obtained from 80 tonnes of iron[ii]oxide

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(Original post by

how do calculate the mass of iron that could be obtained from 80 tonnes of iron[ii]oxide

**Rio ubiteb**)how do calculate the mass of iron that could be obtained from 80 tonnes of iron[ii]oxide

1. Convert 80 tonnes (80000000 g) of Iron (iii) oxide to moles.

2. Ratio is 1:2 so moles of Fe produced is double to moles in 1.

3. Convert moles of Fe in 2 to mass in g and convert to tonnes.

Last edited by username1445490; 2 years ago

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