Question #deae1

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Joshua Z.
Nov 23, 2017

0.0625 moles of lead, which is 12.950 grams of lead

Explanation:

Here is the balanced equation:
Pb_3O4(s)+ 2C(s) --> 3Pb(s) + 2CO_2(g)

The first step is to find the limiting reactant. We can do this by finding the number of moles of each reactant:

molar mass of Pb_3O_4=3*207.2+4*16=685.6 g/(mol)
moles of Pb_3O_4=(40g)/(685.6 g/(mol))=0.0583 mols

molar mass of C=12 g/(mol)
moles of C=(1.0g)/(12 g/(mol))=0.0833 mols
It takes two moles of carbon to create this reaction, so we will divide the moles of C by 2:
(0.0833 mols)/2=0.0417mols

Since 0.0583 mols>0.0417mols, we know that C is the limiting reactant. Now, we can get the number of moles of Pb by multiplying the number of moles of C by the conversion factor. Since 2 moles of carbon makes 3 moles of lead, we have:

(0.0417molsC)*((3molsPb)/(2molsC))=0.0625molsPb

If you want the mass of lead, we can just divide by the molar mass of lead (207.2g/(mol))

0.0625molsPb*207.2g/(mol)=12.950g

If you have any questions or confusions, just comment. Hope this helps!

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