Li4(CO3)2+ 2 H2O +2 CO2 → 4 LiHCO3
What is the mass in grams produced of LiHCO3?
The initial amounts of reactants are still:
Li4(CO3)2 (444 grams),
H2O (200 grams) and
CO2 (352 grams).
Do not change anything about the question ! please calculate it like that
Aw, C'mon lee. Although I changed it, I showed exactly what I did so you could follow the same process and work it any way you wish. You should have learned how to do it from my work.
Here is the process AGAIN.
Li4(CO3)2+ 2 H2O +2 CO2 → 4 LiHCO3
mols Li4(CO3)2 = grams/molar mass = ? You can do the math.
mols H2O = grams/molar mass = ?
mols CO2 = grams/molar mass = ?
I did those earlier and there will be NO change in the number of mols. The values of 3 mols Li4(CO3)2, 11 mols for H2O and 8 mols for CO2 did not change. Now, using the coefficients in the balanced equation convert those numbers to mols LiHCO3.
3 mols Li4(CO3)2 x (4 mols LiHCO3/1 mol) = 12 mols LiHCO3
11 mols H2O x (4 mols LiHCO3/2 mols H2O) = 22 mols LiHCO3
8 mols CO2 x (4 mols LiHCO3/2 mols CO2) = 16 mols LiHCO3.
The limiting reagent STILL is Li4(CO3)2 (although that compound doesn't exist) just as in the previous solution.
Then grams LiHCO3 = mols LiHCO3 x molar mass LiHCO3 = ?
The bottom line, and you could have calculated that yourself, is that you doubled the starting material from Li2CO3 to Li4(CO3)2 so you double the grams of LiHCO3 produced. I hope this helps but I can't help but be curious about how/why you're using Li4(CO3)2.
To calculate the mass of LiHCO3 produced, we need to determine the limiting reactant, which is the reactant that restricts the amount of LiHCO3 that can be formed.
1. Calculate the molar mass of each reactant:
- Li4(CO3)2: 4 lithium (4 x 6.94 g/mol) + 2 carbonate (2 x (12.01 g/mol + 3 x 16.00 g/mol)) = 29.75 g/mol
- H2O: 2 hydrogen (2 x 1.01 g/mol) + 1 oxygen (16.00 g/mol) = 18.02 g/mol
- CO2: 1 carbon (12.01 g/mol) + 2 oxygen (2 x 16.00 g/mol) = 44.01 g/mol
2. Convert the initial amounts of reactants to moles:
- Li4(CO3)2: 444 g / 29.75 g/mol = 14.89 mol
- H2O: 200 g / 18.02 g/mol = 11.10 mol
- CO2: 352 g / 44.01 g/mol = 7.99 mol
3. Determine the stoichiometric ratio between the reactants and the product:
From the balanced equation, we can see that the ratio of Li4(CO3)2:H2O:CO2:LiHCO3 is 1:2:2:4.
4. Calculate the amount of LiHCO3 that can be formed:
- Assuming Li4(CO3)2 is the limiting reactant, we use the stoichiometric ratio to calculate the moles of LiHCO3 produced: 14.89 mol Li4(CO3)2 x (4 mol LiHCO3 / 1 mol Li4(CO3)2) = 59.56 mol LiHCO3
5. Convert the moles of LiHCO3 to grams:
- The molar mass of LiHCO3 is 90.01 g/mol.
- The mass of LiHCO3 produced is: 59.56 mol x 90.01 g/mol = 5,364.45 g
Therefore, the mass of LiHCO3 produced is approximately 5364.45 grams.
To calculate the mass in grams produced of LiHCO3, we will use stoichiometry to determine the amount of LiHCO3 produced from the given reactant amounts.
First, let's determine the molar masses of the reactants and the product:
- Li4(CO3)2 (lithium carbonate): 4 lithium (4 x 6.94 g/mol) + 2 carbon (2 x 12.01 g/mol) + 6 oxygen (6 x 16.00 g/mol) = 73.89 g/mol
- H2O (water): 2 hydrogens (2 x 1.01 g/mol) + 1 oxygen (1 x 16.00 g/mol) = 18.02 g/mol
- CO2 (carbon dioxide): 1 carbon (1 x 12.01 g/mol) + 2 oxygen (2 x 16.00 g/mol) = 44.01 g/mol
- LiHCO3 (lithium bicarbonate): 1 lithium (1 x 6.94 g/mol) + 1 hydrogen (1 x 1.01 g/mol) + 1 carbon (1 x 12.01 g/mol) + 3 oxygen (3 x 16.00 g/mol) = 90.01 g/mol
Next, we need to find the limiting reagent. The limiting reagent is the reactant that is completely consumed and determines the maximum amount of product that can be formed.
To find the limiting reagent, we compare the moles of each reactant to their respective stoichiometric coefficients in the balanced equation.
For Li4(CO3)2:
- Moles of Li4(CO3)2 = mass of Li4(CO3)2 / molar mass of Li4(CO3)2
- Moles of Li4(CO3)2 = 444 g / 73.89 g/mol
- Moles of Li4(CO3)2 ≈ 6.00 mol
For H2O:
- Moles of H2O = mass of H2O / molar mass of H2O
- Moles of H2O = 200 g / 18.02 g/mol
- Moles of H2O ≈ 11.10 mol
For CO2:
- Moles of CO2 = mass of CO2 / molar mass of CO2
- Moles of CO2 = 352 g / 44.01 g/mol
- Moles of CO2 ≈ 8.00 mol
Based on the stoichiometry of the balanced equation, 1 mole of Li4(CO3)2 produces 4 moles of LiHCO3. Therefore, we can calculate the maximum moles of LiHCO3 that can be produced for each reactant:
- Maximum moles of LiHCO3 from Li4(CO3)2 = moles of Li4(CO3)2 x (4 moles LiHCO3 / 1 mole Li4(CO3)2)
- Maximum moles of LiHCO3 from Li4(CO3)2 = 6.00 mol x (4 mol LiHCO3 / 1 mol Li4(CO3)2) = 24.00 mol
- Maximum moles of LiHCO3 from H2O = moles of H2O x (4 moles LiHCO3 / 2 moles H2O)
- Maximum moles of LiHCO3 from H2O = 11.10 mol x (4 mol LiHCO3 / 2 mol H2O) = 22.20 mol
- Maximum moles of LiHCO3 from CO2 = moles of CO2 x (4 moles LiHCO3 / 2 moles CO2)
- Maximum moles of LiHCO3 from CO2 = 8.00 mol x (4 mol LiHCO3 / 2 mol CO2) = 16.00 mol
Based on these calculations, we can see that the limiting reagent is CO2 since it produces the least amount of LiHCO3 (16.00 mol). Therefore, the maximum moles of LiHCO3 that can be produced is 16.00 mol.
Finally, we can calculate the mass of LiHCO3 produced using the limiting reagent:
- Mass of LiHCO3 = moles of LiHCO3 x molar mass of LiHCO3
- Mass of LiHCO3 = 16.00 mol x 90.01 g/mol
- Mass of LiHCO3 ≈ 1,440.16 g
Therefore, the mass in grams produced of LiHCO3 is approximately 1,440.16 grams.