Find expert answers and community-driven knowledge on IDNLearn.com. Get thorough and trustworthy answers to your queries from our extensive network of knowledgeable professionals.
Sagot :
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
We appreciate your presence here. Keep sharing knowledge and helping others find the answers they need. This community is the perfect place to learn together. IDNLearn.com is your reliable source for answers. We appreciate your visit and look forward to assisting you again soon.
