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Certainly! Let's tackle this problem step-by-step to find the weight of [tex]\( \text{CaCO}_3 \)[/tex] needed to produce sufficient [tex]\( \text{CO}_2 \)[/tex] to convert 21.2 kg of [tex]\( \text{Na}_2\text{CO}_3 \)[/tex] to [tex]\( \text{NaHCO}_3 \)[/tex].
### Step 1: Known Data and Molar Masses
To start, let's note down the molar masses involved:
- Molar mass of [tex]\( \text{CaCO}_3 \)[/tex]: 100 g/mol
- Molar mass of [tex]\( \text{CO}_2 \)[/tex]: 44 g/mol
- Molar mass of [tex]\( \text{Na}_2\text{CO}_3 \)[/tex]: 106 g/mol
We are given 21.2 kg (or 21200 grams) of [tex]\( \text{Na}_2\text{CO}_3 \)[/tex].
### Step 2: Convert Mass of [tex]\( \text{Na}_2\text{CO}_3 \)[/tex] to Moles
We need to first find out how many moles of [tex]\( \text{Na}_2\text{CO}_3 \)[/tex] this corresponds to:
[tex]\[ \text{Moles of } \text{Na}_2\text{CO}_3 = \frac{\text{Mass of } \text{Na}_2\text{CO}_3}{\text{Molar mass of } \text{Na}_2\text{CO}_3} \][/tex]
[tex]\[ \text{Moles of } \text{Na}_2\text{CO}_3 = \frac{21200 \text{ g}}{106 \text{ g/mol}} = 200 \text{ moles} \][/tex]
### Step 3: Stoichiometry to Find Moles of [tex]\( \text{CO}_2 \)[/tex] Needed
From the reaction [tex]\( \text{Na}_2\text{CO}_3 + \text{CO}_2 + \text{H}_2\text{O} \rightarrow 2 \text{NaHCO}_3 \)[/tex], we see that 1 mole of [tex]\( \text{Na}_2\text{CO}_3 \)[/tex] reacts with 1 mole of [tex]\( \text{CO}_2 \)[/tex].
Therefore, the moles of [tex]\( \text{CO}_2 \)[/tex] required will also be 200.0 moles.
### Step 4: Convert Moles of [tex]\( \text{CO}_2 \)[/tex] to Mass
Next, we need to convert these moles of [tex]\( \text{CO}_2 \)[/tex] into mass:
[tex]\[ \text{Mass of } \text{CO}_2 = \text{Moles of } \text{CO}_2 \times \text{Molar mass of } \text{CO}_2 \][/tex]
[tex]\[ \text{Mass of } \text{CO}_2 = 200.0 \text{ moles} \times 44 \text{ g/mol} = 8800 \text{ grams} \][/tex]
### Step 5: Stoichiometry to Find Moles of [tex]\( \text{CaCO}_3 \)[/tex] Needed
From the reaction [tex]\( \text{CaCO}_3 \rightarrow \text{CaO} + \text{CO}_2 \)[/tex], we see that 1 mole of [tex]\( \text{CaCO}_3 \)[/tex] produces 1 mole of [tex]\( \text{CO}_2 \)[/tex].
Therefore, the moles of [tex]\( \text{CaCO}_3 \)[/tex] required will also be 200.0 moles.
### Step 6: Convert Moles of [tex]\( \text{CaCO}_3 \)[/tex] to Mass
Finally, we convert these moles of [tex]\( \text{CaCO}_3 \)[/tex] into mass:
[tex]\[ \text{Mass of } \text{CaCO}_3 = \text{Moles of } \text{CaCO}_3 \times \text{Molar mass of } \text{CaCO}_3 \][/tex]
[tex]\[ \text{Mass of } \text{CaCO}_3 = 200.0 \text{ moles} \times 100 \text{ g/mol} = 20000 \text{ grams} \][/tex]
### Conclusion
To produce sufficient [tex]\( \text{CO}_2 \)[/tex] to convert 21.2 kg of [tex]\( \text{Na}_2\text{CO}_3 \)[/tex] to [tex]\( \text{NaHCO}_3 \)[/tex], you need to decompose 20,000 grams (or 20 kg) of [tex]\( \text{CaCO}_3 \)[/tex].
So, the weight of [tex]\( \text{CaCO}_3 \)[/tex] that needs to be decomposed is 20,000 grams (or 20 kg).
### Step 1: Known Data and Molar Masses
To start, let's note down the molar masses involved:
- Molar mass of [tex]\( \text{CaCO}_3 \)[/tex]: 100 g/mol
- Molar mass of [tex]\( \text{CO}_2 \)[/tex]: 44 g/mol
- Molar mass of [tex]\( \text{Na}_2\text{CO}_3 \)[/tex]: 106 g/mol
We are given 21.2 kg (or 21200 grams) of [tex]\( \text{Na}_2\text{CO}_3 \)[/tex].
### Step 2: Convert Mass of [tex]\( \text{Na}_2\text{CO}_3 \)[/tex] to Moles
We need to first find out how many moles of [tex]\( \text{Na}_2\text{CO}_3 \)[/tex] this corresponds to:
[tex]\[ \text{Moles of } \text{Na}_2\text{CO}_3 = \frac{\text{Mass of } \text{Na}_2\text{CO}_3}{\text{Molar mass of } \text{Na}_2\text{CO}_3} \][/tex]
[tex]\[ \text{Moles of } \text{Na}_2\text{CO}_3 = \frac{21200 \text{ g}}{106 \text{ g/mol}} = 200 \text{ moles} \][/tex]
### Step 3: Stoichiometry to Find Moles of [tex]\( \text{CO}_2 \)[/tex] Needed
From the reaction [tex]\( \text{Na}_2\text{CO}_3 + \text{CO}_2 + \text{H}_2\text{O} \rightarrow 2 \text{NaHCO}_3 \)[/tex], we see that 1 mole of [tex]\( \text{Na}_2\text{CO}_3 \)[/tex] reacts with 1 mole of [tex]\( \text{CO}_2 \)[/tex].
Therefore, the moles of [tex]\( \text{CO}_2 \)[/tex] required will also be 200.0 moles.
### Step 4: Convert Moles of [tex]\( \text{CO}_2 \)[/tex] to Mass
Next, we need to convert these moles of [tex]\( \text{CO}_2 \)[/tex] into mass:
[tex]\[ \text{Mass of } \text{CO}_2 = \text{Moles of } \text{CO}_2 \times \text{Molar mass of } \text{CO}_2 \][/tex]
[tex]\[ \text{Mass of } \text{CO}_2 = 200.0 \text{ moles} \times 44 \text{ g/mol} = 8800 \text{ grams} \][/tex]
### Step 5: Stoichiometry to Find Moles of [tex]\( \text{CaCO}_3 \)[/tex] Needed
From the reaction [tex]\( \text{CaCO}_3 \rightarrow \text{CaO} + \text{CO}_2 \)[/tex], we see that 1 mole of [tex]\( \text{CaCO}_3 \)[/tex] produces 1 mole of [tex]\( \text{CO}_2 \)[/tex].
Therefore, the moles of [tex]\( \text{CaCO}_3 \)[/tex] required will also be 200.0 moles.
### Step 6: Convert Moles of [tex]\( \text{CaCO}_3 \)[/tex] to Mass
Finally, we convert these moles of [tex]\( \text{CaCO}_3 \)[/tex] into mass:
[tex]\[ \text{Mass of } \text{CaCO}_3 = \text{Moles of } \text{CaCO}_3 \times \text{Molar mass of } \text{CaCO}_3 \][/tex]
[tex]\[ \text{Mass of } \text{CaCO}_3 = 200.0 \text{ moles} \times 100 \text{ g/mol} = 20000 \text{ grams} \][/tex]
### Conclusion
To produce sufficient [tex]\( \text{CO}_2 \)[/tex] to convert 21.2 kg of [tex]\( \text{Na}_2\text{CO}_3 \)[/tex] to [tex]\( \text{NaHCO}_3 \)[/tex], you need to decompose 20,000 grams (or 20 kg) of [tex]\( \text{CaCO}_3 \)[/tex].
So, the weight of [tex]\( \text{CaCO}_3 \)[/tex] that needs to be decomposed is 20,000 grams (or 20 kg).
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