Calculate Grams of Carbon Residue from Propane

[KYJelly]

An incomplete reaction of 12.125 g of propane proceeds as follows:

3 C3H8(g) + 14 O2(g) → 8 CO2(g) + 12 H2O(g) + C(s)

How many grams of solid carbon residue are produced as a product?

So, C3H8

C=3(12.01)=36.03 g/mol.
H=8(1.01) = 8.08 g/mol.
= 44.11 g/mol.

Therefore, 12.125 g C3H8 x (1 mol. C3H8/44.11g C3H8) = 0.2749 mol C3H8

Then, 0.2749 mol. C3H8 x (1 mol. C/3 mol. C3H8) = 0.09163 mol C

So, 0.09163 mol. C x (12.01 g C/ 1 mol. C) = 1.101 g C?

 

[epenguin]

An incomplete reaction of 12.125 g of propane proceeds as follows:

3 C3H8(g) + 14 O2(g) → 8 CO2(g) + 12 H2O(g) + C(s)

How many grams of solid carbon residue are produced as a product?

So, C3H8

C=3(12.01)=36.03 g/mol.
H=8(1.01) = 8.08 g/mol.
= 44.11 g/mol.

Therefore, 12.125 g C3H8 x (1 mol. C3H8/44.11g C3H8) = 0.2749 mol C3H8

Then, 0.2749 mol. C3H8 x (1 mol. C/3 mol. C3H8) = 0.09163 mol C

So, 0.09163 mol. C x (12.01 g C/ 1 mol. C) = 1.101 g C?

I get the same answer as you but couldn't follow your argument and suggest you won't be able to if you come back to it after 6 months. If people can't trace a simple logical process, you won't get the credit if you make a mistake, nor will you so easily detect a mistake.

I would just set it out 3 moles propane gives 1 mole C;

Therefore (multiplying by molecular masses) so many g propane gives so many g C

Therefore (simple proportions) the stated number of propane gives so many g C

 

[KYJelly]

Thanks epenguin! I'll definitely revamp my form.