Potassium chlorate decomposes according to the following chemical equation:
2KCIO3 -- 2KCI + 3O2
If you start with 458g of KCIO3 , what mass (g) of O2 will be produced?
83.71 L O2
179.39 L O2
687 L O2
67.2 L O2

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

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)

first, consult an activity series chart to determine if al is more reactive than fe. it is, so it will replace the fe. now, one can continue with the problem.

next, make sure you have a correctly balanced equation.

3feo + 2al ==> 3fe + al2o3

now, determine which reactant (feo or al) is in limiting supply.

moles feo present = 125 g x 1 mole/71.8 g = 1.74 moles

moles al present = 25.0 g x 1 mole/26.98 g = 0.927 moles

since from the balanced equation, it take 3 mol feo for every 2 mol al, al will be limiting.

you can determine this simply by dividing moles of each by their coefficient in the balanced equation. for example: 1.74/3 = 0.58 and 0.927/2 = 0.464. al is less, so it is limiting.

finally, using the stoichiometry in the balanced eq. determine mass of fe that can be produced from 0.927 moles of al.

0.927 moles al x 3 moles fe/2 moles al x 55.8 g/mole = 77.6 g

c- cold periods in earth's history during which glaciers covered much of the surface.

explanation: