2pb(s) + o2(aq) + 4h+(aq) → 2h2o(l) + 2pb2+(aq)
if 100 l of 0.13m lead solution is produced from the corrosion reaction, how many grams of lead metal reacted?

the answer is 0.269 g, but i don't understand the math behind getting this answer. ?

The answer to your question is 0.269 g of Pb

Explanation:

Data

Lead solution = 0.000013 M

Volume = 100 L

mass = 0.269 g

atomic mass Pb = 207.2 g

Chemical reaction

                        2Pb(s) + O₂(aq) + 4H⁺(aq) → 2H₂O(l) + 2Pb₂⁺(aq)

Process

1.- Calculate the mass of Pb in solution

Formula

Molarity =

Solve for number of moles

Number of moles = Volume x Molarity

Substitution

Number of moles = 100 x 0.000013

Number of moles = 0.0013

2.- Calculate the mass of Pb formed.

                       207.2 g of Pb ----------------- 1 mol

                             x g             ----------------- 0.0013 moles

                        x = (0.0013 x 207.2) / 1

                        x = 0.269 g of Pb                                                                

0.269 grams of lead metal reacted

Explanation:

Step 1: Data given

Volume of a 0.000013M lead solution = 100 L

Molar mass of Pb = 207.2 g/mol

Step 2: The balanced equation

2Pb(s) + O2(aq) + 4H+(aq) → 2H2O(l) + 2Pb^2+(aq)

Step 3: Calculate moles of lead

Moles Pb^2+ = molarity * volume

Moles Pb^2+ = 0.000013M * 100L

Moles Pb^2+ = 0.00130 moles

Step 4: Calculate moles Pb metal

For 2 moles Pb we need 1 mol O2 and 4 moles H+ to produce 2 moles H2O and 2 moles Pb^2+

For 0.00130 moles Pb^2+ produced we need 0.00130 moles Pb

Step 5: Calculate mass of Pb

Mass Pb = moles Pb * molar mass Pb

Mass Pb = 0.00130 moles * 207.2 g/mol

Mass Pb =  0.269 grams

0.269 grams of lead metal reacted

i think it is true since you can test the answer and can use observations to complete it.