To determine how many grams of [tex]\( O_2(g) \)[/tex] can be produced from heating 98.0 g of [tex]\( KClO_3(s) \)[/tex], we need to follow these steps:
1. Calculate the molar mass of [tex]\( KClO_3 \)[/tex]:
The molar masses of the elements (in g/mol) are:
- Potassium (K): 39.1
- Chlorine (Cl): 35.45
- Oxygen (O): 16.00
Thus, the molar mass of [tex]\( KClO_3 \)[/tex] is:
[tex]\[
\text{Molar mass of } KClO_3 = 39.1 + 35.45 + 3 \times 16.00 = 122.55 \, \text{g/mol}
\][/tex]
2. Calculate the number of moles of [tex]\( KClO_3 \)[/tex] used:
Using the formula:
[tex]\[
\text{Moles of } KClO_3 = \frac{\text{mass of } KClO_3}{\text{molar mass of } KClO_3}
\][/tex]
Given the mass of [tex]\( KClO_3 \)[/tex] is 98.0 g:
[tex]\[
\text{Moles of } KClO_3 = \frac{98.0 \, \text{g}}{122.55 \, \text{g/mol}} \approx 0.79967 \, \text{mol}
\][/tex]
3. Use the stoichiometry of the reaction to find moles of [tex]\( O_2 \)[/tex] produced:
According to the balanced chemical equation:
[tex]\[
2 \, KClO_3 \rightarrow 2 \, KCl + 3 \, O_2
\][/tex]
The ratio of [tex]\( KClO_3 \)[/tex] to [tex]\( O_2 \)[/tex] is 2:3. Therefore, the moles of [tex]\( O_2 \)[/tex] produced can be calculated using the stoichiometric ratio:
[tex]\[
\text{Moles of } O_2 = \left( \frac{3}{2} \right) \times \text{moles of } KClO_3
\][/tex]
[tex]\[
\text{Moles of } O_2 = \left( \frac{3}{2} \right) \times 0.79967 \, \text{mol} \approx 1.19951 \, \text{mol}
\][/tex]
4. Calculate the mass of [tex]\( O_2 \)[/tex] produced:
The molar mass of [tex]\( O_2 \)[/tex] is:
[tex]\[
\text{Molar mass of } O_2 = 2 \times 16.00 = 32.00 \, \text{g/mol}
\][/tex]
Thus, the mass of [tex]\( O_2 \)[/tex] can be determined using:
[tex]\[
\text{Mass of } O_2 = \text{moles of } O_2 \times \text{molar mass of } O_2
\][/tex]
[tex]\[
\text{Mass of } O_2 = 1.19951 \, \text{mol} \times 32.00 \, \text{g/mol} \approx 38.38 \, \text{g}
\][/tex]
Therefore, the mass of [tex]\( O_2 \)[/tex] produced from heating 98.0 g of [tex]\( KClO_3 \)[/tex] is approximately [tex]\(\boxed{38.38 \, \text{g}}\)[/tex].
