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Assoli18 [71]
4 years ago
11

2KCIO3 --- 2 KCI + 3O2

Chemistry
2 answers:
iragen [17]4 years ago
8 0

Answer:

We have to start with 465.69 grams of KClO3 (option 4 is correct)

Explanation:

Step 1: Data given

Moles of O2 = 5.7 moles

Molar mass KClO3 = 122.55 g/mol

Step 2: The balanced equation

2KCIO3 → 2KCI + 3O2

Step 3: Calculate moles KClO3

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

For 5.7 moles O2 we nee 2/3 * 5.7 = 3.8 moles KClO3

Step 4: Calculate mass of KClO3

Mass KClO3 = moles KClO3 * molar mass KClO3

Mass KClO3 = 3.8 moles * 122.55 g/mol

Mass KClO3 = 465.69 grams

We have to start with 465.69 grams of KClO3 (option 4 is correct)

Juli2301 [7.4K]4 years ago
3 0

Answer:

465.69g KCIO3

Explanation:

See the stoichiometry in the reaction:

We can propose

3 moles of oxgen are made of 2 moles of chlorate

Therefore 5.70 moles of O₂ will be made by (5.70. 2) / 3 = 3.8 moles of chlorate.

We convert the moles to mass: 3.8 mol . 122.55 g/ 1 mol = 465.69g KCIO3

is the mass to use in the begining

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Pb(SO4)2 + 4 LiNO3 → Pb(NO3)4 + 2 Li2SO4
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Answer:

4.5 moles of lithium sulfate are produced.

Explanation:

Given data:

Number of moles of lead sulfate = 2.25 mol

Number of moles of lithium nitrate = 9.62 mol

Number of moles of lithium sulfate = ?

Solution:

Chemical equation:

Pb(SO₄)₂ + 4LiNO₃      →     Pb(NO₃)₄ + 2Li₂SO₄

Now we will compare the moles of lithium sulfate with lead sulfate and lithium nitrate.

                       Pb(SO₄)₂        :         Li₂SO₄

                            1                :             2

                          2.25           :          2/1×2.25 = 4.5 mol

                       LiNO₃            :             Li₂SO₄

                           4                :                2

                           9.62           :             2/4×9.62 = 4.81 mol

Pb(SO₄)₂  produces less number of moles of Li₂SO₄ thus it will act as limiting reactant and limit the yield of  Li₂SO₄.      

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Answer: 104 g

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