How many grams of lithium hydroxide are needed to remove 920 g of carbon dioxide in the air

first you need to write out the balanced reaction equation of LiOH reacting with CO2.

So...

2LiOH + CO2--->Li2CO3 + H2O

You start with 920 g CO2 and use this equation to convert to moles CO2 and then to moles LiOH. There are 2 moles LiOH for every 1 mole of CO2. And finally convert to grams of LiOH.

Paul gave it to you straight.

To determine the number of grams of lithium hydroxide needed to remove carbon dioxide, we need to use stoichiometry.

1. Write the balanced chemical equation for the reaction between lithium hydroxide (LiOH) and carbon dioxide (CO2):
2 LiOH + CO2 -> Li2CO3 + H2O

2. Determine the molar mass of LiOH and CO2:
- Molar mass of LiOH = 6.94 g/mol (Lithium) + 15.999 g/mol (Oxygen) + 1.008 g/mol (Hydrogen) = 23.998 g/mol
- Molar mass of CO2 = 12.011 g/mol (Carbon) + (2 * 15.999 g/mol) (Oxygen) = 44.01 g/mol

3. Use the molar masses to perform a conversion from grams to moles for CO2:
Moles of CO2 = Mass of CO2 / Molar mass of CO2
= 920 g / 44.01 g/mol

4. Apply the stoichiometry of the balanced equation to determine the moles of LiOH required:
From the balanced equation, we see that 2 moles of LiOH react with 1 mole of CO2.
Therefore, Moles of LiOH = (2/1) * Moles of CO2

5. Convert moles of LiOH to grams:
Mass of LiOH = Moles of LiOH * Molar mass of LiOH

Now, let's calculate the answer:
1. Calculate the moles of CO2:
Moles of CO2 = 920 g / 44.01 g/mol = 20.89 mol

2. Calculate the moles of LiOH:
Moles of LiOH = (2/1) * 20.89 mol = 41.78 mol

3. Calculate the mass of LiOH:
Mass of LiOH = 41.78 mol * 23.998 g/mol = 1002.83 g

Therefore, approximately 1002.83 grams of lithium hydroxide are needed to remove 920 grams of carbon dioxide from the air.