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Small quantities of oxygen can be prepared in the laboratory by heating potassium chlorate, [tex]\(KClO_3 (s)\)[/tex]. The equation for the reaction is:

[tex]\[
2 KClO_3 \longrightarrow 2 KCl + 3 O_2
\][/tex]

Calculate how many grams of [tex]\(O_2(g)\)[/tex] can be produced from heating [tex]\(58.1 \, \text{g} \, KClO_3(s)\)[/tex].

Mass: [tex]\(\square\)[/tex]

TOOLS: [tex]\(x \times 10^y\)[/tex]


Sagot :

To determine the mass of oxygen ([tex]\(O_2\)[/tex]) produced from heating 58.1 grams of potassium chlorate ([tex]\(KClO_3\)[/tex]), we need to follow a series of steps involving stoichiometric calculations. Here is the detailed, step-by-step solution:

1. Determine the molar mass of [tex]\(KClO_3\)[/tex]:
- The molar mass of [tex]\(KClO_3\)[/tex] is 122.55 g/mol.

2. Calculate the moles of [tex]\(KClO_3\)[/tex]:
[tex]\[ \text{moles of } KClO_3 = \frac{\text{mass of } KClO_3}{\text{molar mass of } KClO_3} = \frac{58.1 \text{ g}}{122.55 \text{ g/mol}} \approx 0.474 \text{ moles} \][/tex]

3. Using the balanced chemical equation, determine the relation between moles of [tex]\(KClO_3\)[/tex] and [tex]\(O_2\)[/tex]:
- According to the equation [tex]\(2 KClO_3 \rightarrow 2 KCl + 3 O_2\)[/tex], 2 moles of [tex]\(KClO_3\)[/tex] produce 3 moles of [tex]\(O_2\)[/tex].

4. Calculate the moles of [tex]\(O_2\)[/tex] produced:
[tex]\[ \text{moles of } O_2 = \left(\frac{3}{2}\right) \times \text{moles of } KClO_3 = \left(\frac{3}{2}\right) \times 0.474 \approx 0.711 \text{ moles} \][/tex]

5. Determine the molar mass of [tex]\(O_2\)[/tex]:
- The molar mass of [tex]\(O_2\)[/tex] is 32.00 g/mol.

6. Calculate the mass of [tex]\(O_2\)[/tex] produced:
[tex]\[ \text{mass of } O_2 = \text{moles of } O_2 \times \text{molar mass of } O_2 = 0.711 \text{ moles} \times 32.00 \text{ g/mol} \approx 22.76 \text{ g} \][/tex]

Therefore, from heating 58.1 grams of [tex]\(KClO_3\)[/tex], approximately 22.76 grams of oxygen ([tex]\(O_2\)[/tex]) can be produced.