Given the following equation, how many grams of PbCO3 will dissolve when exactly 1.0 L of 1.00 M H+ is added to 6.00 g of PbCO3?
When the equation is PbCO3(s) + 2H(+)(aq) = Pb(2+)(aq) + H2O(l) + CO2(g)
Please explain in detail, step by step
4.6
To determine how many grams of PbCO3 will dissolve when 1.0 L of 1.00 M H+ is added to 6.00 g of PbCO3, we need to follow these steps:
Step 1: Calculate the number of moles of H+ present in 1.0 L of 1.00 M H+ solution.
First, we need to calculate the number of moles of H+ in the solution. The concentration is given as 1.00 M, which means there is 1 mole of H+ per liter of solution. Therefore, 1.0 L of 1.00 M H+ solution contains 1.00 moles of H+.
Step 2: Calculate the molar mass of PbCO3.
To calculate the molar mass of PbCO3, we need to determine the atomic masses of each element in the compound.
The atomic masses are:
Pb: 207.2 g/mol
C: 12.01 g/mol
O: 16.00 g/mol
The molar mass of PbCO3 is calculated as follows:
Molar mass of PbCO3 = (207.2 g/mol) + (12.01 g/mol) + (3 × 16.00 g/mol)
= 207.2 g/mol + 12.01 g/mol + 48.00 g/mol
= 267.21 g/mol
Step 3: Convert the given mass of PbCO3 (6.00 g) to moles.
To convert grams to moles, we divide the given mass (in grams) by the molar mass (in grams/mol).
Number of moles of PbCO3 = 6.00 g / 267.21 g/mol
= 0.0225 mol
Step 4: Determine the limiting reactant.
To determine the limiting reactant, we compare the stoichiometric ratio between PbCO3 and H+ in the balanced equation.
From the equation, we can see that two moles of H+ are required to react with one mole of PbCO3. Therefore, the stoichiometric ratio is 2:1.
Since we have 1.00 mole of H+ and 0.0225 moles of PbCO3, we can conclude that PbCO3 is the limiting reactant because it is in smaller quantity.
Step 5: Calculate the moles of PbCO3 that will react.
Since PbCO3 is the limiting reactant, all of it will react with the H+ ions.
Therefore, the moles of PbCO3 that will react are equal to the initial moles of PbCO3, which is 0.0225 mol.
Step 6: Convert moles of PbCO3 to grams.
To convert moles of PbCO3 to grams, we multiply the moles by the molar mass of PbCO3.
Mass of PbCO3 that will react = 0.0225 mol × 267.21 g/mol
= 6.01 g (rounded to two decimal places)
Therefore, when exactly 1.0 L of 1.00 M H+ is added to 6.00 g of PbCO3, approximately 6.01 grams of PbCO3 will dissolve.
To solve this problem, we need to use stoichiometry, which is the relationship between the amounts of reactants and products in a chemical reaction. The equation given allows us to determine the stoichiometric ratio between PbCO3 and H+. Here's how you can solve the problem step by step:
1. Determine the molar mass of PbCO3:
- The molar mass of Pb is 207.2 g/mol.
- The molar mass of C is 12.0 g/mol.
- The molar mass of O is 16.0 g/mol.
- Therefore, the molar mass of PbCO3 is 207.2 + 12.0 + (16.0 * 3) = 267.2 g/mol.
2. Convert the given amount of PbCO3 from grams to moles:
- Divide the mass of PbCO3 (6.00 g) by its molar mass (267.2 g/mol) to get moles of PbCO3.
- Moles of PbCO3 = 6.00 g / 267.2 g/mol = 0.02245 mol.
3. Use the balanced equation to determine the stoichiometric ratio between PbCO3 and H+.
- From the equation, the stoichiometric ratio between PbCO3 and H+ is 1:2, which means it takes 1 mole of PbCO3 to react with 2 moles of H+.
4. Determine the number of moles of H+:
- The concentration of H+ is given as 1.00 M (moles per liter).
- Since we have 1.0 L of H+, the number of moles of H+ is 1.00 M * 1.0 L = 1.00 mol.
5. Use the stoichiometric ratio to determine the number of moles of PbCO3 that will react with H+.
- Since the stoichiometric ratio is 1:2 and we have 1.00 mol of H+, we can conclude that 0.50 mol of PbCO3 will react.
6. Compare the moles of PbCO3 available with the moles of PbCO3 that will react.
- We have 0.02245 mol of PbCO3, and 0.50 mol of PbCO3 will react.
- Since 0.02245 mol is less than 0.50 mol, we conclude that all of the PbCO3 will react.
7. Convert the moles of PbCO3 that will react to grams:
- Multiply the number of moles of PbCO3 (0.50 mol) by its molar mass (267.2 g/mol) to get the mass of PbCO3 that will dissolve.
- Mass of PbCO3 = 0.50 mol * 267.2 g/mol = 133.6 g.
Therefore, when exactly 1.0 L of 1.00 M H+ is added to 6.00 g of PbCO3, the entire 6.00 g of PbCO3 will dissolve.
See the equation in the problem.
How many moles H^+ have you added? That is M H^+ x L H^+ = moles H^+ = 1.00M x 1 L= 1.0 mol H^+.
Using the coefficients in the balanced equation, convert moles H^+ to moles PbCO3.
mols PbCO3 = moles H^+ x (1 mole PbCO3/2 mol H^+) = moles H^+ x 1/2 = 1 x 1/2 = 0.5 mol PbCO3.
Now convert mol PbCO3 to grams. g = moles x molar mass
Here is a worked example that will solve all of your stoichiometry problems. Print it out and keep it.
http://www.jiskha.com/science/chemistry/stoichiometry.html