Question

(Extra points+Brainliest Answer, :) happy holidays)

Answer 1

the answer is 360 g H2O with a 90.3 yeild percentage

Answer 2

Answer : The theoretical yield of water in this reaction is 360 g.

Solution : Given,

Mass of $LiOH=1\times 10^3g=1000g$

Mass of $CO_2=8.80\times 10^2g=880g$

Mass of $H_2O=3.25\times 10^2g=325g$

Molar mass of $CO_2$ = 44 g/mole

Molar mass of $LiOH$ = 24 g/mole

Molar mass of $H_2O$ = 18 g/mole

First we have to calculate the moles of $CO_2$ and $LiOH$.

Moles of $CO_2$ = $\frac{\text{ Mass of }CO_2}{\text{ Molar mass of }CO_2}= \frac{880g}{44g/mole}=20moles$

Moles of $LiOH$ = $\frac{\text{ Mass of }LiOH}{\text{ Molar mass of }LiOH}= \frac{1000g}{24g/mole}=42moles$

The given balanced chemical reaction is,

$CO_2+2LiOH\rightarrow Li_2CO_3+H_2O$

From the given reaction, we conclude that

1 mole of $CO_2$ react with 2 mole of $LiOH$

20 moles of $CO_2$ requires $20\times2=40moles$ of $LiOH$

But the actual moles of $LiOH$ = 42 moles

Excess moles of $LiOH$ = 42 - 40 = 2 moles

So, $CO_2$ is the limiting reagent.

Now we have to calculate the mass of $H_2O$.

From the reaction, we conclude that

1 mole of $CO_2$ gives 1 mole of $H_2O$

20 moles of $CO_2$ gives 20 moles of $H_2O$

Mass of $H_2O$ = Moles of $H_2O$ × Molar mass of  $H_2O$

Mass of $H_2O$ = 20 moles × 18 g/mole = 360 g

The experimental yield of $H_2O$ = 325 g

Therefore, the theoretical yield of $H_2O$ = 360 g