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How many grams of potassium chlorate are required to form 638 ml o2 at 128°c and 752 torr according to the equation below? 2kclo3(s) => 2kcl(s) + 3o2(g)?

Respuesta :

malika8888
2KClO3(s) => 2KCl(s) + 3O2(g)
245 g. ➡️ 3 mol
x. ➡️ 0.02 mol

[tex]x = \frac{245 \times 0.02}{3} \\ x = 1.6 \: g[/tex]





[tex]PV=nRT \\ 1 \times 0.638 = n \times 0.082 \times 401 \\ 0.638 = 32,882n \\ n = 0.02 \: mol[/tex]

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