Welcome to Westonci.ca, the Q&A platform where your questions are met with detailed answers from experienced experts. Connect with professionals ready to provide precise answers to your questions on our comprehensive Q&A platform. Join our Q&A platform to connect with experts dedicated to providing accurate answers to your questions in various fields.
Sagot :
Let's break down the problem step by step to find out how many sodium atoms must react to give us 33.6 dm³ of hydrogen gas at STP, and then calculate the mass of KClO[tex]\(_3\)[/tex] needed to produce 3.36 dm³ of oxygen gas at STP.
### Part 1: Sodium Atoms for Hydrogen Gas Production
Given the reaction:
[tex]\[ 2 Na + 2 H_2O \rightarrow 2 NaOH + H_2(g) \][/tex]
We need to determine how many moles of hydrogen gas are in 33.6 dm³ at STP.
1. Calculate the moles of hydrogen gas (H[tex]\(_2\)[/tex]):
The molar volume of a gas at STP (Standard Temperature and Pressure) is 22.4 dm³/mol. Therefore, the number of moles of H[tex]\(_2\)[/tex] produced can be calculated as:
[tex]\[ \text{Moles of H}_2 = \frac{\text{Volume of H}_2}{\text{Molar Volume}} = \frac{33.6 \, \text{dm}^3}{22.4 \, \text{dm}^3/\text{mol}} = 1.5 \, \text{moles} \][/tex]
2. Determine the moles of sodium (Na) reacting:
According to the balanced equation, 2 moles of sodium produce 1 mole of hydrogen gas. Hence, to produce 1.5 moles of H[tex]\(_2\)[/tex], we need:
[tex]\[ \text{Moles of Na} = 2 \times \text{Moles of H}_2 = 2 \times 1.5 = 3 \, \text{moles} \][/tex]
3. Calculate the number of sodium atoms:
The number of atoms is given by multiplying the moles by Avogadro's number ([tex]\(6.022 \times 10^{23}\)[/tex] atoms/mol):
[tex]\[ \text{Atoms of Na} = \text{Moles of Na} \times \text{Avogadro's number} = 3 \, \text{moles} \times 6.022 \times 10^{23} \, \text{atoms/mol} \approx 1.8066 \times 10^{24} \, \text{atoms} \][/tex]
### Part 2: Mass of KClO[tex]\(_3\)[/tex] for Oxygen Gas Production
Given the reaction:
[tex]\[ 2 KClO_3(s) \rightarrow 2 KCl(s) + 3 O_2(g) \][/tex]
We need to calculate how many moles of oxygen gas are in 3.36 dm³ at STP, and consequently determine the required mass of KClO[tex]\(_3\)[/tex].
1. Calculate the moles of oxygen gas (O[tex]\(_2\)[/tex]):
[tex]\[ \text{Moles of O}_2 = \frac{\text{Volume of O}_2}{\text{Molar Volume}} = \frac{3.36 \, \text{dm}^3}{22.4 \, \text{dm}^3/\text{mol}} = 0.15 \, \text{moles} \][/tex]
2. Determine the moles of KClO[tex]\(_3\)[/tex] required:
According to the balanced equation, 2 moles of KClO[tex]\(_3\)[/tex] decompose to produce 3 moles of O[tex]\(_2\)[/tex]. Therefore:
[tex]\[ \text{Moles of KClO}_3 = \frac{2}{3} \times \text{Moles of O}_2 = \frac{2}{3} \times 0.15 = 0.1 \, \text{moles} \][/tex]
3. Calculate the mass of KClO[tex]\(_3\)[/tex] needed:
The molar mass of KClO[tex]\(_3\)[/tex] is calculated as follows:
[tex]\[ \text{Molar mass of KClO}_3 = 39.1 \, (\text{K}) + 35.5 \, (\text{Cl}) + 3 \times 16 \, (\text{O}) = 122.6 \, \text{g/mol} \][/tex]
Then, the mass of KClO[tex]\(_3\)[/tex] required is:
[tex]\[ \text{Mass of KClO}_3 = \text{Moles of KClO}_3 \times \text{Molar Mass of KClO}_3 = 0.1 \, \text{moles} \times 122.6 \, \text{g/mol} = 12.26 \, \text{g} \][/tex]
### Summary and Final Results
1. To produce 33.6 dm³ of hydrogen gas at STP, [tex]\(1.8066 \times 10^{24}\)[/tex] sodium atoms must react completely.
2. To produce 3.36 dm³ of oxygen gas at STP, 12.26 grams of KClO[tex]\(_3\)[/tex] must decompose completely.
### Part 1: Sodium Atoms for Hydrogen Gas Production
Given the reaction:
[tex]\[ 2 Na + 2 H_2O \rightarrow 2 NaOH + H_2(g) \][/tex]
We need to determine how many moles of hydrogen gas are in 33.6 dm³ at STP.
1. Calculate the moles of hydrogen gas (H[tex]\(_2\)[/tex]):
The molar volume of a gas at STP (Standard Temperature and Pressure) is 22.4 dm³/mol. Therefore, the number of moles of H[tex]\(_2\)[/tex] produced can be calculated as:
[tex]\[ \text{Moles of H}_2 = \frac{\text{Volume of H}_2}{\text{Molar Volume}} = \frac{33.6 \, \text{dm}^3}{22.4 \, \text{dm}^3/\text{mol}} = 1.5 \, \text{moles} \][/tex]
2. Determine the moles of sodium (Na) reacting:
According to the balanced equation, 2 moles of sodium produce 1 mole of hydrogen gas. Hence, to produce 1.5 moles of H[tex]\(_2\)[/tex], we need:
[tex]\[ \text{Moles of Na} = 2 \times \text{Moles of H}_2 = 2 \times 1.5 = 3 \, \text{moles} \][/tex]
3. Calculate the number of sodium atoms:
The number of atoms is given by multiplying the moles by Avogadro's number ([tex]\(6.022 \times 10^{23}\)[/tex] atoms/mol):
[tex]\[ \text{Atoms of Na} = \text{Moles of Na} \times \text{Avogadro's number} = 3 \, \text{moles} \times 6.022 \times 10^{23} \, \text{atoms/mol} \approx 1.8066 \times 10^{24} \, \text{atoms} \][/tex]
### Part 2: Mass of KClO[tex]\(_3\)[/tex] for Oxygen Gas Production
Given the reaction:
[tex]\[ 2 KClO_3(s) \rightarrow 2 KCl(s) + 3 O_2(g) \][/tex]
We need to calculate how many moles of oxygen gas are in 3.36 dm³ at STP, and consequently determine the required mass of KClO[tex]\(_3\)[/tex].
1. Calculate the moles of oxygen gas (O[tex]\(_2\)[/tex]):
[tex]\[ \text{Moles of O}_2 = \frac{\text{Volume of O}_2}{\text{Molar Volume}} = \frac{3.36 \, \text{dm}^3}{22.4 \, \text{dm}^3/\text{mol}} = 0.15 \, \text{moles} \][/tex]
2. Determine the moles of KClO[tex]\(_3\)[/tex] required:
According to the balanced equation, 2 moles of KClO[tex]\(_3\)[/tex] decompose to produce 3 moles of O[tex]\(_2\)[/tex]. Therefore:
[tex]\[ \text{Moles of KClO}_3 = \frac{2}{3} \times \text{Moles of O}_2 = \frac{2}{3} \times 0.15 = 0.1 \, \text{moles} \][/tex]
3. Calculate the mass of KClO[tex]\(_3\)[/tex] needed:
The molar mass of KClO[tex]\(_3\)[/tex] is calculated as follows:
[tex]\[ \text{Molar mass of KClO}_3 = 39.1 \, (\text{K}) + 35.5 \, (\text{Cl}) + 3 \times 16 \, (\text{O}) = 122.6 \, \text{g/mol} \][/tex]
Then, the mass of KClO[tex]\(_3\)[/tex] required is:
[tex]\[ \text{Mass of KClO}_3 = \text{Moles of KClO}_3 \times \text{Molar Mass of KClO}_3 = 0.1 \, \text{moles} \times 122.6 \, \text{g/mol} = 12.26 \, \text{g} \][/tex]
### Summary and Final Results
1. To produce 33.6 dm³ of hydrogen gas at STP, [tex]\(1.8066 \times 10^{24}\)[/tex] sodium atoms must react completely.
2. To produce 3.36 dm³ of oxygen gas at STP, 12.26 grams of KClO[tex]\(_3\)[/tex] must decompose completely.
Thanks for using our service. We aim to provide the most accurate answers for all your queries. Visit us again for more insights. We hope this was helpful. Please come back whenever you need more information or answers to your queries. Discover more at Westonci.ca. Return for the latest expert answers and updates on various topics.
