Question

If this reaction produced 51.5 g of KCl, how much O2 was produced (in grams)?

Answer 1

Question #47825 – Chemistry – Other

Question:

If this reaction produced 51.5 g of KCl, how much O₂ was produced (in grams)?

Answer:

The reaction equation is:

$$2 \mathrm{KClO_3} = 2 \mathrm{KCl} + 3 \mathrm{O_2}$$

The amount of moles of KCl is:

$$n(\mathrm{KCl}) = \frac{m(\mathrm{KCl})}{M(\mathrm{KCl})} = \frac{51.5\,g}{74.5\,g/mol} = 0.69\,\mathrm{moles}$$

The amount of moles O₂ produced is:

$$n(\mathrm{O_2}) = \frac{n(\mathrm{KCl}) \times 3}{2} = \frac{0.69 \times 3}{2} = 1.035\,\mathrm{moles}$$

Than the mass of O₂ is:

$$m(\mathrm{O_2}) = n(\mathrm{O_2}) \times M(\mathrm{O_2}) = 1.035\,\mathrm{moles} \times 32\,g/mol = 33.12\,g$$

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