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ahrayia [7]
3 years ago
8

Potassium chlorate decomposes according to the following chemical equation:

Chemistry
2 answers:
kari74 [83]3 years ago
5 0

Answer:

179.4 g of oyxgen will be produced.

Option 2.

Be careful because the units says L instead of g

Explanation:

Decomposition reaction:

2KClO₃ → 2KCl + 3O₂

2 moles of chlorate can decompose to 2 moles of chloride and 3 moles of oxygen.

We convert the mass to moles, and then we make the rule of three:

458 g / 122.55 g/mol =3.74 moles of chlorate

2 moles of chlorate can produce 3 moles of oxygen

Then, 3.74 moles of chlorate may produce (3.74 .3) / 2 = 5.60 moles of O₂

We convert the moles to mass → 5.60 mol . 32 g / 1mol = 179.4 g

Sedaia [141]3 years ago
5 0

Answer:

179.39 grams of O2 will be produced (option 2 is correct)

Explanation:

Step 1: Data given

Mass of KClO3 = 458 grams

Molar mass KClO3 = 122.55 g/mol

Molar mass O2 = 32.0 g/mol

Step 2: The balanced equation

2KCIO3 → 2KCI + 3O2

Step 3: Calculate moles KClO3

Moles KClO3 = mass KClO3 / molar mass KClO3

Moles KClO3 = 458 grams / 122.55 g/mol

Moles KClO3 = 3.74 moles

Step 4: Calculate moles of O2

For 2 moles KClO3 we'll have 2 moles KCl and 3 moles O2

For 3.74 moles we'll have 3/2 * 3.74 = 5.606 moles

Step 5: Calculate mass O2

Mass O2 = 5.606 moles * 32.0 g/mol

Mass O2 = 179.39 grams

179.39 grams of O2 will be produced (option 2 is correct)

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aseous methane will react with gaseous oxygen to produce gaseous carbon dioxide and gaseous water . Suppose 5.5 g of methane is
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Answer:

There is 9.6 grams of CO2 produced

Explanation:

Step 1: Data given

Mass of methane = 5.50 grams

Molar mass of methane = 16.04 g/mol

Mass of oxygen = 13.9 grams

Molar mass of oxygen = 32.0 g/mol

Step 2: The reaction

CH4(g) + 2O2(g) → CO2(g) + 2H2O(g)

Step 3: Calculate number of moles

Moles = mass / molar mass

Moles methane = 5.50 grams / 16.04 g/mol

Moles methane = 0.343 moles

Moles oxygen = 13.9 grams / 32.0 g/mol

Moles oxygen = 0.434 moles

For 1 mol CH4 we need 2 moles O2 to produce 1 mol CO2 and 2 moles H2O

O2 is the limiting reactant. It will completely react (0.434 moles).

There will react 0.434/2 = 0.217 moles CH4

There will remain 0.343-0.217 = 0.126 moles CH4

There will be produced 0.434 moles of H2O and

0.434/2 =0.217 moles of CO2

Step 4: Calculate mass of products

Mass = moles * molar mass

Mass CO2 = 0.217 moles ¨44.01 g/mol

Mass CO2 = 9.6 grams

Mass H2O = 0.434 moles * 18.02

Mass H2O = 7.8 grams

4 0
3 years ago
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