How many grams of PbCO3 will dissolve in 1.00L of 1.00M H+ is added to 5.00g of PbCO3??

PbCO3(s) + 2H+ (aq) -> Pb2+ (aq) + H2O (l) + CO2 (g)

According to my calculations all of it will dissolve. Here is a worked example of a stoichiometry problem. Follow the steps. I thin about 130 grams or so should dissolve and you don't have that much.

http://www.jiskha.com/science/chemistry/stoichiometry.html

To find out how many grams of PbCO3 will dissolve in 1.00L of 1.00M H+, we need to determine which reactant is limiting and calculate the amount of product formed using stoichiometry.

Here's how you can proceed:

1. Determine the moles of H+ in 1.00L of 1.00M H+:
Moles of H+ = Molarity × Volume
= 1.00 M × 1.00 L
= 1.00 mol

2. Calculate the moles of PbCO3 using the given mass:
Moles of PbCO3 = Mass / Molar mass
= 5.00 g / (207.2 g/mol + 12.0 g/mol + 48.0 g/mol)
= 0.02384 mol (rounded to 4 decimal places)

3. According to the balanced equation, the stoichiometric ratio between PbCO3 and H+ is 1:2. This means that for every 1 mole of PbCO3, 2 moles of H+ react.

4. Let's compare the moles of PbCO3 to the available moles of H+:

H+ moles / PbCO3 moles = 1.00 mol / 0.02384 mol
= 41.9 (rounded to 1 decimal place)

5. Based on the stoichiometric ratio, this means that PbCO3 is in excess, and all 0.02384 moles will react.

6. Finally, calculate the mass of PbCO3 that will dissolve:
Mass of PbCO3 = Moles of PbCO3 × Molar mass
= 0.02384 mol × (207.2 g/mol + 12.0 g/mol + 48.0 g/mol)
= 5.06 g (rounded to 2 decimal places)

Therefore, approximately 5.06 grams of PbCO3 will dissolve in 1.00L of 1.00M H+ when 5.00g of PbCO3 is added.