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vlada-n [284]
3 years ago
14

How many moles of sodium are needed to produce 2.4 mol of sodium oxide?

Chemistry
1 answer:
Lorico [155]3 years ago
3 0

Answer:

Explanation:

4

N

a

+

O

2

→

2

N

a

2

O

.

By the stoichiometry of this reaction if 5 mol natrium react, then 2.5 mol

N

a

2

O

should result.

Explanation:

The molecular mass of natrium oxide is

61.98

g

⋅

m

o

l

−

1

. If

5

m

o

l

natrium react, then

5

2

m

o

l

×

61.98

g

⋅

m

o

l

−

1

=

154.95

g

natrium oxide should result.

So what have I done here? First, I had a balanced chemical equation (this is the important step; is it balanced?). Then I used the stoichiometry to get the molar quantity of product, and converted this molar quantity to mass. If this is not clear, I am willing to have another go.

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Answer:

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Sodium metal and water react to create sodium hydrogen and hydrogen gas through the unbalanced equation.
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Answer:

Theoretical yield = 2.5 g

Explanation:

Given data:

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Solution:

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2Na + 2H₂O → 2NaOH + H₂

Number of moles of sodium:

Number of moles = mass/ molar mass

Number of moles = 79.7 g / 23 g/mol

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Number of moles of water:

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Number of moles = 2.5 mol

Now we will compare the moles of hydrogen gas with water and sodium.

                        H₂O           :             H₂

                           2             :              1

                          2.5           :            1/2×2.5 =1.25 mol

                     

                           Na           :              H₂

                             2            :               1

                           3.5           :             1/2×3.5 =1.75 mol

water will be limiting reactant.

Theoretical yield:

Mass = number of moles × molar mass

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Before we can use this equation for
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Answer:

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Explanation:

Chemical equation:

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Left hand side                      Right hand side

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Left hand side                      Right hand side

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H = 12                                    H = 2

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Left hand side                      Right hand side

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O = 2                                     O = 14

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