To determine how much lime ([tex]$\text{CaO}$[/tex]) is produced when heating 50 kg of limestone ([tex]$\text{CaCO}_3$[/tex]) to [tex]$825^\circ C$[/tex], given that 22 kg of carbon dioxide ([tex]$\text{CO}_2$[/tex]) is produced, follow these steps:
1. Write the balanced chemical equation:
[tex]\[
\text{CaCO}_3 \rightarrow \text{CaO} + \text{CO}_2
\][/tex]
2. Understand the molar relationship:
The balanced equation tells us that 1 mole of [tex]$\text{CaCO}_3$[/tex] decomposes to produce 1 mole of [tex]$\text{CaO}$[/tex] and 1 mole of [tex]$\text{CO}_2$[/tex].
3. Determine the molar masses:
- Molar mass of [tex]$\text{CaCO}_3$[/tex] (limestone): 100 g/mol
- Molar mass of [tex]$\text{CaO}$[/tex] (lime): 56 g/mol
- Molar mass of [tex]$\text{CO}_2$[/tex]: 44 g/mol
4. Use the given data:
From the problem, 50 kg (or 50,000 g) of limestone produces 22 kg (or 22,000 g) of [tex]$\text{CO}_2$[/tex].
5. Relate the masses using stoichiometry:
For every 44 g of [tex]$\text{CO}_2$[/tex] produced, 56 g of [tex]$\text{CaO}$[/tex] is produced.
6. Scale up to the given amount of [tex]$\text{CO}_2$[/tex]:
If 22,000 g of [tex]$\text{CO}_2$[/tex] is produced, we can set up a proportion to find the mass of [tex]$\text{CaO}$[/tex] produced.
[tex]\[
\frac{56 \text{ g CaO}}{44 \text{ g CO}_2} = \frac{x \text{ g CaO}}{22,000 \text{ g CO}_2}
\][/tex]
7. Solve for [tex]\( x \)[/tex]:
[tex]\[
x = \frac{56 \ \text{g CaO}}{44 \ \text{g CO}_2} \times 22,000 \ \text{g CO}_2
\][/tex]
Simplifying the right-hand side:
[tex]\[
x = \frac{56}{44} \times 22,000
\][/tex]
[tex]\[
x = 1.2727 \times 22,000
\][/tex]
[tex]\[
x = 28,000 \ \text{g}
\][/tex]
Since 1 kg = 1,000 g:
[tex]\[
x = 28,000 \ \text{g} = 28 \ \text{kg}
\][/tex]
So, when 50 kg of limestone is heated, it produces 28 kg of lime ([tex]$\text{CaO}$[/tex]).
The correct answer is:
B 28 kg
