Why am I getting the answers wrong when the method is right?

This is on Problems on reacting masses of solids

What mass of glucose must be fermented to give 5.0kg of ethanol?
(don't know how to get subscript)

C6H12(aq)----> 2C2H5OH (aq)+ 2CO2

Ok... so I'm trying to find the moles in ethanol

To convert 5.0Kg of ethanol to grammes: X1000 which gives 5000g
Find the RMM of ethanol: (12*2)+(1*5)+ 16+ 1= 46

To find the moles: mass/Rmm which is 5000/46= 108.7 mols
Ratio of C6H12O6: C2H5O8
1 : 2

1/2*108.7= 54.35 mol

Rmm of glucose= (12*6)+(1*12)+(16*6)= 180

To find the mass of glucose: RMM* MOLES
Which is 180*54.35= 9783g

The answer they gave us on the sheet was 9782.64g.

I know they're essentially the same answer but I've done the same working for other questions and got a completely different answer.... I stupidly typed up the one equation I got right... I'll type up another one later. I just don't know why I'm getting the answer marginally wrong when I haven't rounded before the final answer....
Can someone please point out where I'm going wrong? Thanks!


Edit: and also can someone show me how to work through this question? It's an extension but I really want to know how to do it...

A mixture of calcium and magnesium carbonates weighing 10.00g was heated until it reached a constant mass of 5.0960g. Calculate the percentage composition of the mixture by mass.
CaCo3(s)----> CaO + Co2 :: MgCO3 -----> MgO+ CO2

I have the answers on the sheet (CaCO3= 39.86%, MgCO3= 60.14%) but I have no idea how to get to the answer as we haven't been taught it and my textbook is useless.

Thank you!!!!

This is on Problems on reacting masses of solids

What mass of glucose must be fermented to give 5.0kg of ethanol?
(don't know how to get subscript)

C6H12(aq)----> 2C2H5OH (aq)+ 2CO2

Ok... so I'm trying to find the moles in ethanol

To convert 5.0Kg of ethanol to grammes: X1000 which gives 5000g
Find the RMM of ethanol: (12*2)+(1*5)+ 16+ 1= 46

To find the moles: mass/Rmm which is 5000/46= 108.7 mols
Ratio of C6H12O6: C2H5O8
1 : 2

1/2*108.7= 54.35 mol

Rmm of glucose= (12*6)+(1*12)+(16*6)= 180

To find the mass of glucose: RMM* MOLES
Which is 180*54.35= 9783g

The answer they gave us on the sheet was 9782.64g.

I know they're essentially the same answer but I've done the same working for other questions and got a completely different answer.... I stupidly typed up the one equation I got right... I'll type up another one later. I just don't know why I'm getting the answer marginally wrong when I haven't rounded before the final answer....
Can someone please point out where I'm going wrong? Thanks!


Edit: and also can someone show me how to work through this question? It's an extension but I really want to know how to do it...

A mixture of calcium and magnesium carbonates weighing 10.00g was heated until it reached a constant mass of 5.0960g. Calculate the percentage composition of the mixture by mass.
CaCo3(s)----> CaO + Co2 :: MgCO3 -----> MgO+ CO2

I have the answers on the sheet (CaCO3= 39.86%, MgCO3= 60.14%) but I have no idea how to get to the answer as we haven't been taught it and my textbook is useless.

Thank you!!!!

Due to the theory of conservation of mass, then what you started with = what you ended with(the total mass), so we know that carbon dioxide escaped.

Therefore if you do a balanced eqn with MgCO3 and CaCO3 decomposed by heat, you would get 2 moles of CO2 in the end.

The mass of CO2 escaped = 10.00 - 5.0960 = 4.904 g

Mole of the CO2 = 4.904/44 = 0.1114545

Mole of CaCO3 actually present = 1/2 mole of CO2 produced = 0.05572725 mol = mass / 100

therefore mass of CaCO3 actually present = 5.572725 g
Hence the mass of MgCO3 present = 10 - mass of CaCO3 = 4.427275 g

Hmm, seems like i get a different answer, what do you think i have done wrong in my method?

This is your error - you have assumed that there are equal number of moles of the two carbonates

loss of CO2 = 4.904g = 0.11145 moles

let mass of CaCO3 in the mixture = x
therefore mass of MgCO3 in teh mixture = 10-x

therefore moles of CaCO3 in the mix = x/100
therefore moles of MgCO3 in the mix = (10-x)/84
as 1 mole of both produces one mole of CO2 then moles of CO2 = x/100 + (10-x)/84

Hence: x/100 + (10-x)/84 = 0.11145

84x + 1000 - 100x = 936.18

16x = 63.82

x = 3.989 (or roughly 4g)

Therefore mass of CaCO3 = 4g and mass of MgCO3 = 6g

Hiya, I was just wondering if you could explain why the moles of MgCO3 in the mix is shown as (10-x)/84? I understand the 10-x but can't see why it's over 84. Thank you

swear that s the Mr of MgCO3

This thread is 12 years old.

Due to the theory of conservation of mass, then what you started with = what you ended with(the total mass), so we know that carbon dioxide escaped.

Therefore if you do a balanced eqn with MgCO3 and CaCO3 decomposed by heat, you would get 2 moles of CO2 in the end.

The mass of CO2 escaped = 10.00 - 5.0960 = 4.904 g

Mole of the CO2 = 4.904/44 = 0.1114545

Mole of CaCO3 actually present = 1/2 mole of CO2 produced = 0.05572725 mol = mass / 100

therefore mass of CaCO3 actually present = 5.572725 g
Hence the mass of MgCO3 present = 10 - mass of CaCO3 = 4.427275 g

Hmm, seems like i get a different answer, what do you think i have done wrong in my method?

Hiya, I was just wondering if you could explain why the moles of MgCO3 in the mix is shown as (10-x)/84? I understand the 10-x but can't see why it's over 84. Thank you

How did you get 84x 1000 100x

Hence: x/100 + (10-x)/84 = 0.11145

84x + 1000 - 100x = 936.18