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ss7ja [257]
3 years ago
10

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

Chemistry
2 answers:
grigory [225]3 years ago
7 0

the answer is 360 g H2O with a 90.3 yeild percentage

uranmaximum [27]3 years ago
4 0

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

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