How Do You Calculate the Mass of Lead Sulfate Formed in a Lead-Acid Battery?

[whit88]

I'm having a lot of trouble with this problem!

What mass of lead sulfate is formed in a lead-acid storage battery when 1.34 g of Pb undergoes oxidation?

the answer needs to be in grams

 

[Borek]

First you will need at least skeleton reaction equation.

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[Blueshift5]

Balancing redox reactions can be a challenging task, but with practice and understanding of the principles involved, it can become easier. In this specific problem, we are dealing with a lead-acid storage battery. The reaction involved is the oxidation of lead (Pb) to lead sulfate (PbSO4).

To balance this reaction, we need to ensure that the number of atoms of each element is the same on both sides of the equation. In this case, we have one atom of lead on both sides, but the number of sulfur and oxygen atoms differs. To balance this, we can add a coefficient of 1 to PbSO4 on the product side, giving us a balanced equation of Pb + PbSO4 → 2PbSO4.

Now, we can use the balanced equation to determine the mass of lead sulfate formed. From the equation, we can see that 1 mole of Pb reacts with 1 mole of PbSO4. The molar mass of Pb is 207.2 g/mol, so 1.34 g of Pb is equivalent to 0.00646 moles. Therefore, 0.00646 moles of PbSO4 will also be formed.

The molar mass of PbSO4 is 303.3 g/mol, so 0.00646 moles is equivalent to 1.96 grams of PbSO4. Therefore, the mass of lead sulfate formed in this reaction is 1.96 grams.

I hope this explanation helps you understand the process of balancing redox reactions and solving stoichiometry problems. Remember to always start by balancing the equation and then use the mole ratio to determine the mass of the desired product. Keep practicing, and you will become more comfortable with these types of problems.