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).
To determine the mass of LiHCO3 produced, we need to calculate the limiting reactant first. The limiting reactant is the reactant that is completely consumed, limiting the amount of product that can be formed.
Let's calculate how many moles of each reactant are present:
1. Li4(CO3)2:
- Given mass = 444 grams
- Molar mass of Li4(CO3)2 = 4 * (6.94 g/mol) + 2 * (12.01 g/mol) + 6 * (16.00 g/mol)
= 37.88 g/mol
- Moles of Li4(CO3)2 = given mass / molar mass
= 444 g / 37.88 g/mol
= 11.71 mol
2. H2O:
- Given mass = 200 grams
- Molar mass of H2O = 2 * (1.01 g/mol) + 16.00 g/mol
= 18.02 g/mol
- Moles of H2O = given mass / molar mass
= 200 g / 18.02 g/mol
= 11.10 mol
3. CO2:
- Given mass = 352 grams
- Molar mass of CO2 = 12.01 g/mol + 2 * (16.00 g/mol)
= 44.01 g/mol
- Moles of CO2 = given mass / molar mass
= 352 g / 44.01 g/mol
= 7.99 mol
Based on the balanced chemical equation, the stoichiometric ratio between Li4(CO3)2 and LiHCO3 is 1:4. This means that 1 mole of Li4(CO3)2 produces 4 moles of LiHCO3.
Now, let's find the limiting reactant:
1. Determine the ratio of each reactant to the product using the stoichiometric coefficients:
- For Li4(CO3)2: 1 mol Li4(CO3)2 → 4 mol LiHCO3
- For H2O: 1 mol H2O → 4 mol LiHCO3
- For CO2: 1 mol CO2 → 4 mol LiHCO3
2. Multiply the molar ratio by the number of moles for each reactant to calculate the number of moles of LiHCO3 that can be formed:
- Moles of LiHCO3 from Li4(CO3)2 = 11.71 mol Li4(CO3)2 * (4 mol LiHCO3 / 1 mol Li4(CO3)2)
= 46.84 mol LiHCO3
- Moles of LiHCO3 from H2O = 11.10 mol H2O * (4 mol LiHCO3 / 1 mol H2O)
= 44.40 mol LiHCO3
- Moles of LiHCO3 from CO2 = 7.99 mol CO2 * (4 mol LiHCO3 / 1 mol CO2)
= 31.96 mol LiHCO3
The limiting reactant is the reactant that produces the fewest moles of LiHCO3. In this case, CO2 produces the least number of moles of LiHCO3 at 31.96 mol.
Now, let's calculate the mass of LiHCO3 produced using the limiting reactant:
1. Calculate the molar mass of LiHCO3:
- Molar mass of LiHCO3 = (6.94 g/mol) + (1.01 g/mol) + (12.01 g/mol) + 3 * (16.00 g/mol)
= 90.01 g/mol
2. Calculate the mass of LiHCO3 produced using the number of moles from the limiting reactant:
- Mass of LiHCO3 = Moles of LiHCO3 from limiting reactant * molar mass
= 31.96 mol * 90.01 g/mol
= 2877.77 grams
Therefore, the mass of LiHCO3 produced is approximately 2877.77 grams.
Li4(CO3)2+ 2 H2O +2 CO2 → 4 LiHCO3
Does Li4(CO3)2 exist. I don't think so. I will rewrite the equation.
Li2CO3 + H2O + CO2 ==> 2LiHCO3
This is a limiting reagent (LR) problem. First we must determine the LR.
The easy way to do this is to determine the mols of each reactant and convert each to the products. The reactant producing the smallest amount of product is the LR. Note that 444 g Li4(CO3)2 = 222 g Li2CO3.
mols Li2CO3 = grams/molar mass = 222 g/73.8 = 3.0
mols H2O = 200/18 = 11
mols CO2 = 352/44 = 8
Now take these 1 at a time to see mols LiHCO3 formed.
1 mol of EACH will produce 2 mols of the product so.
3 mols of Li2CO3 will produce 6 mols LiHCO3.
11 mols H2O will produce 22 mols LiHCO3.
8 mols CO2 will produce 16 mols LiHCO3.
It is clear that 6 mols LiHCO3 will be produced.
grams LiHCO3 = mols LiHCO3 x molar mass LiHCO3.