2LiOH + Co2 = Li2CO3 +H2O If a cannister absorbed 100.0g of CO2 from the air that resulted in the production of 148.5g of Li2CO3, what was the percent yeild for Li2CO3.

To calculate the percent yield for Li2CO3, we need to compare the actual yield (148.5g) to the theoretical yield, which is the maximum amount that could have been produced according to the balanced equation.

First, let's calculate the moles of CO2 absorbed. We know the molar mass of CO2 is 44.01 g/mol, so we divide the mass (100.0g) by the molar mass to get moles:

moles of CO2 = mass / molar mass
moles of CO2 = 100.0g / 44.01 g/mol
moles of CO2 = 2.27 mol

According to the balanced equation, the mole ratio between CO2 and Li2CO3 is 1:1. Therefore, the moles of Li2CO3 produced should be the same as the moles of CO2:

moles of Li2CO3 = 2.27 mol

Now we need to calculate the theoretical yield of Li2CO3. The molar mass of Li2CO3 is 73.89 g/mol, so we multiply the moles of Li2CO3 by the molar mass to get the theoretical yield in grams:

theoretical yield = moles of Li2CO3 * molar mass
theoretical yield = 2.27 mol * 73.89 g/mol
theoretical yield = 167.67 g

The theoretical yield of Li2CO3 is 167.67g.

Finally, we can calculate the percent yield:

percent yield = (actual yield / theoretical yield) * 100
percent yield = (148.5g / 167.67g) * 100
percent yield = 88.5%

Therefore, the percent yield for Li2CO3 is 88.5%.

To calculate the percent yield of Li2CO3, we need to compare the actual yield of Li2CO3 (148.5g) with the theoretical yield of Li2CO3. The theoretical yield is the maximum amount of product that can be obtained based on stoichiometry.

First, let's determine the balanced chemical equation for the reaction:

2LiOH + CO2 -> Li2CO3 + H2O

From the equation, we can see that the molar ratio between Li2CO3 and CO2 is 1:1. This means that for every mole of CO2, we should obtain an equal amount of moles of Li2CO3.

We need to convert the mass of CO2 absorbed (100.0g) into moles. The molar mass of CO2 is 44.01g/mol.

Moles of CO2 = Mass of CO2 / Molar mass of CO2
= 100.0g / 44.01g/mol
≈ 2.27 mol

Since the molar ratio of Li2CO3 to CO2 is 1:1, the moles of Li2CO3 produced should be the same as the moles of CO2 absorbed.

Therefore, the theoretical yield of Li2CO3 is also 2.27 moles.

Now, let's convert the theoretical moles of Li2CO3 to grams. The molar mass of Li2CO3 is 73.89 g/mol.

Theoretical yield of Li2CO3 = Moles of Li2CO3 * Molar mass of Li2CO3
= 2.27 mol * 73.89 g/mol
≈ 167.62 g

The theoretical yield of Li2CO3 is approximately 167.62 g.

To calculate the percent yield, we use the formula:

Percent Yield = (Actual Yield / Theoretical Yield) * 100

Substituting the values:

Percent Yield = (148.5 g / 167.62 g) * 100
= 88.53%

Therefore, the percent yield of Li2CO3 is approximately 88.53%.