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.
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
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