Question

Question #deae1

Answer 1

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!