Prior to their Phaseout in the 1980s, chemicals containing lead were commonly added to gasoline as anti-knocking agents. A 4.083 sample of one such additive containing only lead, carbon, and hydrogen was burned in an oxygen-rich environment. the products of the combustion were 4.444 g of CO2 and 2.274 g of H2O. Insert subscripts to HCPb.
I found the mole of each (C- .1010, H-.0307, and Pb- 0.01370). After dividing by the lowest number I get H-2.22, C-0.1010, and Pb- 0.01370. I am having trouble finding the correct subscripts.
3 answers
Ms Sue Help me
Help us help you by answering the following.
1. How can the product be ONLY CO2 and H2O. Where is the Pb? And is that Pb, or PbO, PbO2, or Pb3O4? I suspect Pb but I don't know that because you don't list the temperature of the combustion.
2. Post your work for the numbers you have but we must the products first.
However, I will assume Pb is the material.
So 4083 g sample
with 4.444 g CO2
with 2.274 g H2O
with ? g Pb. I get approx 2.6 g Pb but you need a better number than that.
Convert to mols and divide by the smallest number (which will be the mols Pb). Post your work and if you have trouble I can help you through it.
Hint: this is tetraethyl lead.
Do this:
g C = 4.444 x 12/44 = ?
g H = 2.274 x 2/18 = ?
g Pb = 4.083-g C - gH = ?
1. How can the product be ONLY CO2 and H2O. Where is the Pb? And is that Pb, or PbO, PbO2, or Pb3O4? I suspect Pb but I don't know that because you don't list the temperature of the combustion.
2. Post your work for the numbers you have but we must the products first.
However, I will assume Pb is the material.
So 4083 g sample
with 4.444 g CO2
with 2.274 g H2O
with ? g Pb. I get approx 2.6 g Pb but you need a better number than that.
Convert to mols and divide by the smallest number (which will be the mols Pb). Post your work and if you have trouble I can help you through it.
Hint: this is tetraethyl lead.
Do this:
g C = 4.444 x 12/44 = ?
g H = 2.274 x 2/18 = ?
g Pb = 4.083-g C - gH = ?
1. The problem that I listed was the only information supplied- there is no mention of lead as a product, and no combustion temp.
2. g C = (4.444)(12.01)/(44.01) = 1.213 g
g H = (2.274)(2)(1.01) / 18.02 = .255 g
4.083- 1.468 = 2.615 g Pb/207.2 = .01262 mol Pb
.1010 mol C -> 8
.2524 mol H -> 20
.01262 mol Pb -> 1
Is this correct? C8H20Pb
2. g C = (4.444)(12.01)/(44.01) = 1.213 g
g H = (2.274)(2)(1.01) / 18.02 = .255 g
4.083- 1.468 = 2.615 g Pb/207.2 = .01262 mol Pb
.1010 mol C -> 8
.2524 mol H -> 20
.01262 mol Pb -> 1
Is this correct? C8H20Pb
3 answers