Answer :
Sure, let's work through the steps required to solve this problem.
1. Identify the given information:
- Initial amount of [tex]\( \text{KHCO}_3 \)[/tex]: 20 grams
- Molecular weight of [tex]\( \text{KHCO}_3 \)[/tex]: 100 g/mol
- Molecular weight of [tex]\( \text{K}_2 \text{CO}_3 \)[/tex]: 138 g/mol
- Molecular weight of [tex]\( \text{H}_2 \text{O} \)[/tex]: 18 g/mol
- Molecular weight of [tex]\( \text{CO}_2 \)[/tex]: 44 g/mol
2. Calculate the moles of [tex]\( \text{KHCO}_3 \)[/tex] initial amount:
To calculate the moles, we use the formula:
[tex]\[ \text{Moles} = \frac{\text{mass}}{\text{molecular weight}} \][/tex]
Given that the mass of [tex]\( \text{KHCO}_3 \)[/tex] is 20 grams and the molecular weight is 100 g/mol:
[tex]\[ \text{Moles of } \text{KHCO}_3 = \frac{20 \text{ grams}}{100 \text{ g/mol}} = 0.2 \text{ moles} \][/tex]
3. Understand the stoichiometry of the reaction:
The balanced chemical equation is:
[tex]\[ 2 \text{KHCO}_3 \rightarrow \text{K}_2 \text{CO}_3 + \text{H}_2 \text{O} + \text{CO}_2 \][/tex]
This equation tells us that:
- 2 moles of [tex]\( \text{KHCO}_3 \)[/tex] produce 1 mole of [tex]\( \text{K}_2 \text{CO}_3 \)[/tex]
- 2 moles of [tex]\( \text{KHCO}_3 \)[/tex] produce 1 mole of [tex]\( \text{H}_2 \text{O} \)[/tex]
- 2 moles of [tex]\( \text{KHCO}_3 \)[/tex] produce 1 mole of [tex]\( \text{CO}_2 \)[/tex]
Therefore, 1 mole of [tex]\( \text{KHCO}_3 \)[/tex] would produce:
- 0.5 moles of [tex]\( \text{K}_2 \text{CO}_3 \)[/tex]
- 0.5 moles of [tex]\( \text{H}_2 \text{O} \)[/tex]
- 0.5 moles of [tex]\( \text{CO}_2 \)[/tex]
Since we have 0.2 moles of [tex]\( \text{KHCO}_3 \)[/tex]:
- Moles of [tex]\( \text{K}_2 \text{CO}_3 \)[/tex] formed: [tex]\( 0.2 \times 0.5 = 0.1 \)[/tex] moles
- Moles of [tex]\( \text{H}_2 \text{O} \)[/tex] formed: [tex]\( 0.2 \times 0.5 = 0.1 \)[/tex] moles
- Moles of [tex]\( \text{CO}_2 \)[/tex] formed: [tex]\( 0.2 \times 0.5 = 0.1 \)[/tex] moles
4. Calculate the mass of each product:
We use the formula:
[tex]\[ \text{Mass} = \text{moles} \times \text{molecular weight} \][/tex]
- For [tex]\( \text{K}_2 \text{CO}_3 \)[/tex]:
[tex]\[ \text{Mass of } \text{K}_2 \text{CO}_3 = 0.1 \text{ moles} \times 138 \text{ g/mol} = 13.8 \text{ grams} \][/tex]
- For [tex]\( \text{H}_2 \text{O} \)[/tex]:
[tex]\[ \text{Mass of } \text{H}_2 \text{O} = 0.1 \text{ moles} \times 18 \text{ g/mol} = 1.8 \text{ grams} \][/tex]
- For [tex]\( \text{CO}_2 \)[/tex]:
[tex]\[ \text{Mass of } \text{CO}_2 = 0.1 \text{ moles} \times 44 \text{ g/mol} = 4.4 \text{ grams} \][/tex]
Summarizing these results, the masses of the products formed from 20 grams of [tex]\( \text{KHCO}_3 \)[/tex] are:
- [tex]\( \text{K}_2 \text{CO}_3 \)[/tex]: 13.8 grams
- [tex]\( \text{H}_2 \text{O} \)[/tex]: 1.8 grams
- [tex]\( \text{CO}_2 \)[/tex]: 4.4 grams
1. Identify the given information:
- Initial amount of [tex]\( \text{KHCO}_3 \)[/tex]: 20 grams
- Molecular weight of [tex]\( \text{KHCO}_3 \)[/tex]: 100 g/mol
- Molecular weight of [tex]\( \text{K}_2 \text{CO}_3 \)[/tex]: 138 g/mol
- Molecular weight of [tex]\( \text{H}_2 \text{O} \)[/tex]: 18 g/mol
- Molecular weight of [tex]\( \text{CO}_2 \)[/tex]: 44 g/mol
2. Calculate the moles of [tex]\( \text{KHCO}_3 \)[/tex] initial amount:
To calculate the moles, we use the formula:
[tex]\[ \text{Moles} = \frac{\text{mass}}{\text{molecular weight}} \][/tex]
Given that the mass of [tex]\( \text{KHCO}_3 \)[/tex] is 20 grams and the molecular weight is 100 g/mol:
[tex]\[ \text{Moles of } \text{KHCO}_3 = \frac{20 \text{ grams}}{100 \text{ g/mol}} = 0.2 \text{ moles} \][/tex]
3. Understand the stoichiometry of the reaction:
The balanced chemical equation is:
[tex]\[ 2 \text{KHCO}_3 \rightarrow \text{K}_2 \text{CO}_3 + \text{H}_2 \text{O} + \text{CO}_2 \][/tex]
This equation tells us that:
- 2 moles of [tex]\( \text{KHCO}_3 \)[/tex] produce 1 mole of [tex]\( \text{K}_2 \text{CO}_3 \)[/tex]
- 2 moles of [tex]\( \text{KHCO}_3 \)[/tex] produce 1 mole of [tex]\( \text{H}_2 \text{O} \)[/tex]
- 2 moles of [tex]\( \text{KHCO}_3 \)[/tex] produce 1 mole of [tex]\( \text{CO}_2 \)[/tex]
Therefore, 1 mole of [tex]\( \text{KHCO}_3 \)[/tex] would produce:
- 0.5 moles of [tex]\( \text{K}_2 \text{CO}_3 \)[/tex]
- 0.5 moles of [tex]\( \text{H}_2 \text{O} \)[/tex]
- 0.5 moles of [tex]\( \text{CO}_2 \)[/tex]
Since we have 0.2 moles of [tex]\( \text{KHCO}_3 \)[/tex]:
- Moles of [tex]\( \text{K}_2 \text{CO}_3 \)[/tex] formed: [tex]\( 0.2 \times 0.5 = 0.1 \)[/tex] moles
- Moles of [tex]\( \text{H}_2 \text{O} \)[/tex] formed: [tex]\( 0.2 \times 0.5 = 0.1 \)[/tex] moles
- Moles of [tex]\( \text{CO}_2 \)[/tex] formed: [tex]\( 0.2 \times 0.5 = 0.1 \)[/tex] moles
4. Calculate the mass of each product:
We use the formula:
[tex]\[ \text{Mass} = \text{moles} \times \text{molecular weight} \][/tex]
- For [tex]\( \text{K}_2 \text{CO}_3 \)[/tex]:
[tex]\[ \text{Mass of } \text{K}_2 \text{CO}_3 = 0.1 \text{ moles} \times 138 \text{ g/mol} = 13.8 \text{ grams} \][/tex]
- For [tex]\( \text{H}_2 \text{O} \)[/tex]:
[tex]\[ \text{Mass of } \text{H}_2 \text{O} = 0.1 \text{ moles} \times 18 \text{ g/mol} = 1.8 \text{ grams} \][/tex]
- For [tex]\( \text{CO}_2 \)[/tex]:
[tex]\[ \text{Mass of } \text{CO}_2 = 0.1 \text{ moles} \times 44 \text{ g/mol} = 4.4 \text{ grams} \][/tex]
Summarizing these results, the masses of the products formed from 20 grams of [tex]\( \text{KHCO}_3 \)[/tex] are:
- [tex]\( \text{K}_2 \text{CO}_3 \)[/tex]: 13.8 grams
- [tex]\( \text{H}_2 \text{O} \)[/tex]: 1.8 grams
- [tex]\( \text{CO}_2 \)[/tex]: 4.4 grams
