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.
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).
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