Answer :
Sure! Let's work through this problem step-by-step.
The decomposition reaction of potassium chlorate ([tex]\(KClO_3\)[/tex]) is given by the equation:
[tex]\[ 2 KClO_3 \rightarrow 2 KCl + 3 O_2 \][/tex]
This equation shows that 2 moles of [tex]\(KClO_3\)[/tex] decompose to produce 3 moles of [tex]\(O_2\)[/tex].
We are given 45.8 grams of potassium chlorate and need to find out how many grams of oxygen gas ([tex]\(O_2\)[/tex]) can be produced.
1. Calculate the molar mass of potassium chlorate ([tex]\(KClO_3\)[/tex]):
- Potassium (K): 39.10 g/mol
- Chlorine (Cl): 35.45 g/mol
- Oxygen (O): 16.00 g/mol
So, the molar mass of [tex]\(KClO_3\)[/tex] is:
[tex]\[ 39.10 + 35.45 + 3 \times 16.00 = 122.55 \text{ g/mol} \][/tex]
2. Calculate the number of moles of [tex]\(KClO_3\)[/tex] in 45.8 grams:
[tex]\[ \text{Moles of } KClO_3 = \frac{\text{mass}}{\text{molar mass}} = \frac{45.8 \text{ g}}{122.55 \text{ g/mol}} \approx 0.3737 \text{ moles} \][/tex]
3. Determine the moles of [tex]\(O_2\)[/tex] produced:
According to the balanced chemical equation, 2 moles of [tex]\(KClO_3\)[/tex] produce 3 moles of [tex]\(O_2\)[/tex]. Therefore, the moles of [tex]\(O_2\)[/tex] produced by 0.3737 moles of [tex]\(KClO_3\)[/tex] can be calculated as:
[tex]\[ \text{Moles of } O_2 = \left( \frac{3}{2} \right) \times 0.3737 \approx 0.5606 \text{ moles} \][/tex]
4. Calculate the mass of [tex]\(O_2\)[/tex] produced:
The molar mass of [tex]\(O_2\)[/tex] is [tex]\(32.00 \text{ g/mol}\)[/tex].
Therefore, the mass of [tex]\(O_2\)[/tex] produced is:
[tex]\[ \text{Mass of } O_2 = \text{moles} \times \text{molar mass} = 0.5606 \text{ moles} \times 32.00 \text{ g/mol} \approx 17.94 \text{ g} \][/tex]
So, the number of grams of oxygen gas that can be produced from the decomposition of 45.8 grams of potassium chlorate is approximately [tex]\( \boxed{17.9 \text{ g}} \)[/tex].
The decomposition reaction of potassium chlorate ([tex]\(KClO_3\)[/tex]) is given by the equation:
[tex]\[ 2 KClO_3 \rightarrow 2 KCl + 3 O_2 \][/tex]
This equation shows that 2 moles of [tex]\(KClO_3\)[/tex] decompose to produce 3 moles of [tex]\(O_2\)[/tex].
We are given 45.8 grams of potassium chlorate and need to find out how many grams of oxygen gas ([tex]\(O_2\)[/tex]) can be produced.
1. Calculate the molar mass of potassium chlorate ([tex]\(KClO_3\)[/tex]):
- Potassium (K): 39.10 g/mol
- Chlorine (Cl): 35.45 g/mol
- Oxygen (O): 16.00 g/mol
So, the molar mass of [tex]\(KClO_3\)[/tex] is:
[tex]\[ 39.10 + 35.45 + 3 \times 16.00 = 122.55 \text{ g/mol} \][/tex]
2. Calculate the number of moles of [tex]\(KClO_3\)[/tex] in 45.8 grams:
[tex]\[ \text{Moles of } KClO_3 = \frac{\text{mass}}{\text{molar mass}} = \frac{45.8 \text{ g}}{122.55 \text{ g/mol}} \approx 0.3737 \text{ moles} \][/tex]
3. Determine the moles of [tex]\(O_2\)[/tex] produced:
According to the balanced chemical equation, 2 moles of [tex]\(KClO_3\)[/tex] produce 3 moles of [tex]\(O_2\)[/tex]. Therefore, the moles of [tex]\(O_2\)[/tex] produced by 0.3737 moles of [tex]\(KClO_3\)[/tex] can be calculated as:
[tex]\[ \text{Moles of } O_2 = \left( \frac{3}{2} \right) \times 0.3737 \approx 0.5606 \text{ moles} \][/tex]
4. Calculate the mass of [tex]\(O_2\)[/tex] produced:
The molar mass of [tex]\(O_2\)[/tex] is [tex]\(32.00 \text{ g/mol}\)[/tex].
Therefore, the mass of [tex]\(O_2\)[/tex] produced is:
[tex]\[ \text{Mass of } O_2 = \text{moles} \times \text{molar mass} = 0.5606 \text{ moles} \times 32.00 \text{ g/mol} \approx 17.94 \text{ g} \][/tex]
So, the number of grams of oxygen gas that can be produced from the decomposition of 45.8 grams of potassium chlorate is approximately [tex]\( \boxed{17.9 \text{ g}} \)[/tex].