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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.
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