What can i do?plz solve this for me.

Question

what mass of lead sulfite is formed ina a lead acid storage battery when 1g of Pb undergoes oxidation?

Answer 1

ok, i can't give you the complete answer, per se, but listen up:

1) balance the equation

2) find the mole ratios between lead and lead sulfite (PbSO2 - but depends on the charge of lead)

3)the mole ratios are the amount of compounds in a balanced equation

4)using the molar mass (sum of atomic masses), use the formula n=m/M (where n= moles, m= mass (1g) and M=molar mass (of lead)) and find the mole of Pb *do not look at the numbers in front of compounds when calculating molar mass - include only the subscripts

5) set up a mole ratio - EXAMPLE: so if #of mole/1mole = #of mole PbSO2/2 mole, and then use cross-multiplying to get the mole of lead sulfite

6) having the mole of lead sulfite, and with the molar mass (add atomic mass of EACH element - lead + oxygen + sulfur )

7) using the formula n=m/M, rearange for m: m=nM

8) and find the mass

pretty straight forward  - mass of A -> mole of A -> mole of B -> mass B

hope this helped

:D

Answer 2

There is no lead (??) sulfite formed in a lead acid battery.  The anion is sulfate, SO4^2-.

During the discharge of a lead acid battery, lead (IV) oxide gets reduced to PbSO4, and Pb metal gets oxidized to PbSO4.

In charging a lead acid battery, lead (II) disproportionates to form metallic lead and lead (IV) oxide, PbO2.

As to your question you MUST consider the entire redox reaction, not just the oxidation of lead.

Pb(s) + PbO2(s) + 2H2SO4(aq) -->  2PbSO4(s) + 2H2O(l)

1.00 g Pb x (1 mole Pb / 207.2g Pb) x (2 mol PbSO4 / 1 mol Pb) x (303.3 g PbSO4 / 1 mol PbSO4) = 2.93 g PbSO4

====== Follow up =======

"<3" has good intentions but there is no PbSO2.  It is certainly not lead sulfite or lead sulfate.  She also advocates using ratios.  In the 40+ years that I've been teaching chemistry, I have found that solving stoichiometry problems with ratios is often confusing and leads to setting up the wrong ratio.  The unit-factor method, on the other hand, uses the units to define the conversion factor, and there can NEVER be any doubt of the correct setup.