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zmey [24]
3 years ago
11

☆Only need information for problem 3 but you need to know the information from problem 2 to get answer 3☆

Chemistry
1 answer:
svet-max [94.6K]3 years ago
4 0

Answer:

2. The balanced equation is given below

2KClO₃ —> 2KCl + 3O₂

18 moles of oxygen, O₂ were obtained.

3. 21 moles of oxygen, O₂.

Explanation:

2. Determination of the number of mole of oxygen produced.

We'll begin by writing the balanced equation for the reaction. This is illustrated below:

2KClO₃ —> 2KCl + 3O₂

From the balanced equation above,

2 moles of KClO₃ decomposed to produce 3 moles of O₂.

Finally, we shall determine the number of mole oxygen, O₂ produced by the reaction of 12 moles of potassium chlorate, KClO₃. This can be obtained as follow:

From the balanced equation above,

2 moles of KClO₃ decomposed to produce 3 moles of O₂.

Therefore, 12 moles of KClO₃ will decompose to produce = (12 × 3)/2 = 18 moles of O₂.

Thus, 18 moles of oxygen, O₂ were obtained from the reaction

3. Determination of the number of mole of oxygen, O₂ produced by the reaction of 14 moles of KClO₃.

From the balanced equation above,

2 moles of KClO₃ decomposed to produce 3 moles of O₂.

Therefore, 14 moles of KClO₃ will decompose to produce = (14 × 3)/2 = 21 moles of O₂.

Thus, 21 moles of oxygen, O₂ were obtained from the reaction

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A compound is formed when 12.2 g Mg combines completely with 5.16 g N. What is the percent composition of this compound? Please
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3Mg + N₂ → Mg₃N₂
n(Mg) = 12,2g÷24,4g/mol = 0,5mol - limiting reagente.
n(N₂) = 5,16g÷28g/mol = 0,18mol
n(Mg₃N₂):n(Mg) = 1:3, n(Mg₃N₂) = 0,166mol
m(Mg₃N₂) = 0,166mol·101,2g/mol = 16,8g.
%(N)= 2·Ar(N)÷Mr(Mg₃N₂) = 2·14÷101,2 = 27,66% = 0,2766
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              or 100% - 27,66% = 72,34%.
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