Question

If 183.7g of KCO3 is completely burned catalytically. What volume of the oxygen gas will be obtain at 39c at 1200 torr pressure.

Answer 1

Question #72126, Chemistry / Inorganic Chemistry

If 183.7g of KClO₃ is completely burned catalytically. What volume of the oxygen gas will be obtained at 39c at 1200 torr pressure.

Answer:

2KClO₃ = 2KCl + 3O₂

Moles of KClO₃:

$$n(KClO_3) = \frac{m(KClO_3)}{M(KClO_3)} = \frac{183.7\ g}{122.55\ \frac{g}{mol}} = 1.499\ mol$$

According to chemical equation:

$$n(O_2) = \frac{3}{2} \times n(KClO_3) = \frac{3}{2} \times 1.499\ mol = 2.2485\ mol$$

Gas law:

$$\begin{array}{c} pV = nRT \\ p = 1200\ torr = 159986.8\ Pa \\ T = 39\ ^{\circ}\mathrm{C} = 312.15\ K \\ \end{array}$$ $$V = \frac{nRT}{p} = \frac{2.2485\ mol \times 8.31\ \frac{J}{K \times mol} \times 312.15\ K}{159986.8\ Pa} = 0.0365\ m^3 = 36.5\ L$$

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