Respuesta :
Answer : The theoretical yield of water in this reaction is 360 g.
Solution : Given,
Mass of [tex]LiOH=1\times 10^3g=1000g[/tex]
Mass of [tex]CO_2=8.80\times 10^2g=880g[/tex]
Mass of [tex]H_2O=3.25\times 10^2g=325g[/tex]
Molar mass of [tex]CO_2[/tex] = 44 g/mole
Molar mass of [tex]LiOH[/tex] = 24 g/mole
Molar mass of [tex]H_2O[/tex] = 18 g/mole
First we have to calculate the moles of [tex]CO_2[/tex] and [tex]LiOH[/tex].
Moles of [tex]CO_2[/tex] = [tex]\frac{\text{ Mass of }CO_2}{\text{ Molar mass of }CO_2}= \frac{880g}{44g/mole}=20moles[/tex]
Moles of [tex]LiOH[/tex] = [tex]\frac{\text{ Mass of }LiOH}{\text{ Molar mass of }LiOH}= \frac{1000g}{24g/mole}=42moles[/tex]
The given balanced chemical reaction is,
[tex]CO_2+2LiOH\rightarrow Li_2CO_3+H_2O[/tex]
From the given reaction, we conclude that
1 mole of [tex]CO_2[/tex] react with 2 mole of [tex]LiOH[/tex]
20 moles of [tex]CO_2[/tex] requires [tex]20\times2=40moles[/tex] of [tex]LiOH[/tex]
But the actual moles of [tex]LiOH[/tex] = 42 moles
Excess moles of [tex]LiOH[/tex] = 42 - 40 = 2 moles
So, [tex]CO_2[/tex] is the limiting reagent.
Now we have to calculate the mass of [tex]H_2O[/tex].
From the reaction, we conclude that
1 mole of [tex]CO_2[/tex] gives 1 mole of [tex]H_2O[/tex]
20 moles of [tex]CO_2[/tex] gives 20 moles of [tex]H_2O[/tex]
Mass of [tex]H_2O[/tex] = Moles of [tex]H_2O[/tex] × Molar mass of [tex]H_2O[/tex]
Mass of [tex]H_2O[/tex] = 20 moles × 18 g/mole = 360 g
The experimental yield of [tex]H_2O[/tex] = 325 g
Therefore, the theoretical yield of [tex]H_2O[/tex] = 360 g
Answer:
Part 1: limiting reactant is CO2
Part 2: theoretical yield is 360
Part 3: percent yield is 90.3
Explanation:
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