The chemical equation below shows the reaction between carbon dioxide [tex](CO_2)[/tex] and lithium hydroxide [tex](LiOH)[/tex].

[tex]\[ CO_2 + 2LiOH \rightarrow Li_2CO_3 + H_2O \][/tex]

The molar mass of [tex]CO_2[/tex] is [tex]44.01 \, g/mol[/tex]. How many moles of [tex]LiOH[/tex] are needed to react completely with [tex]25.5 \, g[/tex] of [tex]CO_2[/tex]?

A. 0.290 moles
B. 0.579 moles
C. 1.16 moles
D. 1.73 moles



Answer :

To determine how many moles of lithium hydroxide (LiOH) are needed to react completely with 25.5 grams of carbon dioxide (CO₂), we will follow a systematic approach.

### Step 1: Calculate the number of moles of CO₂.
First, let's find the number of moles of CO₂ present in 25.5 grams.

The molar mass of CO₂ is given as 44.01 g/mol. The number of moles \( n \) can be calculated using the formula:
[tex]\[ n = \frac{\text{mass}}{\text{molar mass}} \][/tex]

So,
[tex]\[ n_{\text{CO}_2} = \frac{25.5 \text{ g}}{44.01 \text{ g/mol}} \][/tex]

Solving this gives:
[tex]\[ n_{\text{CO}_2} \approx 0.579 \text{ moles} \][/tex]

### Step 2: Determine the stoichiometric ratio from the balanced equation.
The balanced chemical equation is:
[tex]\[ CO_2 + 2 \, LiOH \rightarrow Li_2CO_3 + H_2O \][/tex]

From the equation, we can see that 1 mole of CO₂ reacts with 2 moles of LiOH.

### Step 3: Calculate the number of moles of LiOH required.
If we have 0.579 moles of CO₂, we need twice that amount of LiOH based on the stoichiometric ratio:
[tex]\[ \text{moles of LiOH} = 2 \times n_{\text{CO}_2} \][/tex]

Substituting the value of \( n_{\text{CO}_2} \):
[tex]\[ \text{moles of LiOH} = 2 \times 0.579 \approx 1.16 \text{ moles} \][/tex]

### Conclusion:
From our calculations, we find that approximately 1.16 moles of LiOH are needed to react completely with 25.5 grams of CO₂. Therefore, the correct option is:

1.16 moles