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angeli.p
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In the manufacture of calcium carbide (CaC2), what is the maximum mass of calcium carbide that can be obtained from 40.0kg of calcium oxide and 4.0kg of carbon?
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Guarddyyy
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Write a balanced symbol equation and determine the moles of both calcium oxide and carbon used.

You now MUST base your calculations on the limiting factor (the one with less moles) and then you can use stoichiometry to calculate the maximum mass of Calcium Carbide.

Now try it and post your calculations here, if you would like me to see if you've done it correctly.
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angeli.p
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equation: CaO+3C----> CaC2+CO

molesCaO= 40000/40.1 +16=713.0mol
molesC= 4000/12=333.34mol

ratio: CaO:CaC2= 1:1 therefore 713.0 mol of CaC2

mass nxMr 713* (40.1+24)= 45703.30 grams

ratio C:CaC2 3:1 333.34/3= 111.11mol of CaC2
mass 111.11* (40.1+24)= 712215.15grams

unsure as to what bit is wrong
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angeli.p
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(Original post by Guarddyyy)
Write a balanced symbol equation and determine the moles of both calcium oxide and carbon used.

You now MUST base your calculations on the limiting factor (the one with less moles) and then you can use stoichiometry to calculate the maximum mass of Calcium Carbide.

Now try it and post your calculations here, if you would like me to see if you've done it correctly.

  1. equation: CaO+3C----> CaC2+CO

    molesCaO= 40000/40.1 +16=713.0mol
    molesC= 4000/12=333.34mol

    ratio: CaO:CaC2= 1:1 therefore 713.0 mol of CaC2

    mass nxMr 713* (40.1+24)= 45703.30 grams

    ratio C:CaC2 3:1 333.34/3= 111.11mol of CaC2
    mass 111.11* (40.1+24)= 712215.15grams

    unsure as to what bit is wrong
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Guarddyyy
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(Original post by angeli.p)
equation: CaO+3C----> CaC2+CO

molesCaO= 40000/40.1 +16=713.0mol
molesC= 4000/12=333.34mol

ratio: CaO:CaC2= 1:1 therefore 713.0 mol of CaC2

mass nxMr 713* (40.1+24)= 45703.30 grams

ratio C:CaC2 3:1 333.34/3= 111.11mol of CaC2
mass 111.11* (40.1+24)= 712215.15grams

unsure as to what bit is wrong
The equation seems correct, CaO + 3C -> CaC2 + CO.

There are fewer moles of carbon in total than calcium oxide, so you must base it off 333.34 moles instead.

The second part seems correct when you used the 111.11 moles, and 713.00 moles would be incorrect, so the maximum amount of calcium carbide you can acquire is 71.22kg (you wrote 712.215kg, so you might and put your decimal in the wrong place).
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Guarddyyy
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Do you have the answer or is this a piece of homework where you will hand it into your teacher to mark?
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angeli.p
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(Original post by Guarddyyy)
Do you have the answer or is this a piece of homework where you will hand it into your teacher to mark?
its hw but how can you get 71kg when theres only like 44kg product
are u supposed to minus it instead
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Guarddyyy
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I'd have to look at this in more detail later, that's actually a very good point... I see why you're confused now.
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Guarddyyy
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TutorsChemistry Could you please take a look at this? Seems like I'm going wrong as well.
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Guarddyyy
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Mr CaC2 = 64.1
Moles of CaC2 = 111.11

111.11*64.1 = 7122.15(grams) = 7.1kg

When we did nMr = m, we came out with a mass of 7122.151, which is 7.1kg, so the answer should be 7.1kg of maximum yield.
Last edited by Guarddyyy; 4 months ago
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