Answer :
To determine how many moles of lithium hydroxide (LiOH) are needed to react completely with 25.5 grams of carbon dioxide (COâ‚‚), we will follow a systematic approach.
### Step 1: Calculate the number of moles of COâ‚‚.
First, let's find the number of moles of COâ‚‚ present in 25.5 grams.
The molar mass of COâ‚‚ 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 COâ‚‚ reacts with 2 moles of LiOH.
### Step 3: Calculate the number of moles of LiOH required.
If we have 0.579 moles of COâ‚‚, 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 COâ‚‚. Therefore, the correct option is:
1.16 moles
### Step 1: Calculate the number of moles of COâ‚‚.
First, let's find the number of moles of COâ‚‚ present in 25.5 grams.
The molar mass of COâ‚‚ 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 COâ‚‚ reacts with 2 moles of LiOH.
### Step 3: Calculate the number of moles of LiOH required.
If we have 0.579 moles of COâ‚‚, 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 COâ‚‚. Therefore, the correct option is:
1.16 moles
