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Macmill

20.8 g 0₂ x
grams of KCIO:
122.55 g KCIO,
32.00 g 0₂
74.55 g KCI
Answer Bank
1 mole KCIO,
1 mole O₂
1 mole KCI
2 moles KCIO,
3 moles 0₂
2 moles KCI
= g KCIO,
g KCIO,

1 Answer

5 votes

Answer:

To solve this problem, we need to use the balanced chemical equation for the reaction between oxygen gas (O₂) and potassium chloride (KCI) to form potassium chlorate (KCIO₃):

2KCI + 3O₂ → 2KCIO₃

We can use the given mass of O₂ (20.8 g) and the molar mass of O₂ (32.00 g/mol) to find the moles of O₂:

20.8 g O₂ x (1 mol O₂ / 32.00 g O₂) = 0.65 mol O₂

According to the balanced equation, 3 moles of O₂ react with 2 moles of KCI to produce 2 moles of KCIO₃. Therefore, we can use the moles of O₂ to find the moles of KCIO₃:

0.65 mol O₂ x (2 mol KCIO₃ / 3 mol O₂) = 0.43 mol KCIO₃

Finally, we can use the molar mass of KCIO₃ (122.55 g/mol) to convert moles of KCIO₃ to grams of KCIO₃:

0.43 mol KCIO₃ x (122.55 g KCIO₃ / 1 mol KCIO₃) = 52.71 g KCIO₃

Therefore, the grams of KCIO₃ that can be produced from 20.8 g O₂ is 52.71 g KCIO₃. However, the problem does not ask for the grams of KCIO₃, but instead asks for the grams of KCIO, which is not a valid compound. It is possible that there is a typo in the problem and that it should have asked for the grams of KCIO₃ instead.

Step-by-step explanation:

User CJR
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