Calculate the grams of NaOH and of LiOH needed to remove 288 g of CO₂ from the air.

What the chemistry is saying

Both sodium hydroxide and lithium hydroxide remove carbon dioxide by making a carbonate and water. The key idea is the mole ratio: each $CO_2$ molecule needs two $OH^-$ (so two moles of $MOH$).

Write and use the balanced equations

For sodium hydroxide: $$2NaOH + CO_2 \rightarrow Na_2CO_3 + H_2O$$ For lithium hydroxide: $$2LiOH + CO_2 \rightarrow Li_2CO_3 + H_2O$$ So, in both cases: $$\frac{\text{mol }MOH}{\text{mol }CO_2} = 2$$

Convert 288 g of CO₂ to moles

Using $M_{CO_2} = 44.01\,\text{g/mol}$: $$n(CO_2)=\frac{288}{44.01}=6.544\,\text{mol}$$

Find moles of hydroxide required

$$n(MOH)=2\, n(CO_2)=2(6.544)=13.088\,\text{mol}$$

Convert moles to grams for NaOH and LiOH

Sodium hydroxide

$M_{NaOH} = 22.99+16.00+1.01 \approx 40.00\,\text{g/mol}$ $$m(NaOH)=13.088\times 40.00=523.5\,\text{g}\approx 5.24\times 10^2\,\text{g}$$

Lithium hydroxide

$M_{LiOH} = 6.94+16.00+1.01 \approx 23.95\,\text{g/mol}$ $$m(LiOH)=13.088\times 23.95=313.5\,\text{g}\approx 3.14\times 10^2\,\text{g}$$

Quick reasonableness check

You need the same number of moles of hydroxide in both cases, but $LiOH$ has a smaller molar mass than $NaOH$, so it should take fewer grams. That matches the results.