Question

How many moles of lithium hydroxide would be required to produce 38.5 g of Li₂CO₃ in the following chemical reaction?

Answer 1

Answer:

Numer of moles of Li₂CO₃:

$\longrightarrow\: n_{(Li_2 CO_3)} = \dfrac{38.5}{(6 \times 2) + 12 + (16 \times 3)}$

$\longrightarrow\: n_{(Li_2 CO_3)} = \dfrac{38.5}{12+ 12 +48}$

$\longrightarrow\: n_{(Li_2 CO_3)} = \dfrac{38.5}{72}$

$\longrightarrow\: n_{(Li_2 CO_3)} = 0.53473 \: mol$

Chemical Reaction:

2 LiOH(s) + CO₂(g) → Li₂CO₃(s) + H₂O(l)

• 1 mole of Li₂CO₃ is formed from 2 moles of LiOH.

Therefore, applying unitary method:

• 0.53473 mole of Li₂CO₃ is formed from 2 × 0.53473 = 1.06946 moles of LiOH.