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mafiozo [28]
3 years ago
12

How many moles of hydrogen are needed to produce 13.78 mol of ethane?

Chemistry
2 answers:
mamaluj [8]3 years ago
4 0
Ethane is an alkane. Methane is also an alkane and is considered to be the simplest alkane. The difference is ethane has only 2 carbon. That carbon has 6 hydrogen attached to it. So what we do is we multiply the moles of ethane by the number of hydrogen (by dimension analysis) resulting to 82.68 moles H.
Likurg_2 [28]3 years ago
4 0

This is an incomplete question, here is a complete question.

The following balanced equation shows the formation of ethane (C₂H₆).

C_2H_2+2H_2\rightarrow C_2H_6

How many moles of hydrogen are needed to produce 13.78 mol of ethane?

1) 3.445 mol

2) 6.890 mol

3) 27.56 mol

4) 55.12 mol

Answer : The number of moles of hydrogen needed are, (3) 27.56 mol

Explanation : Given,

Moles of ethane = 13.78 mol

The given balanced chemical reaction is:

C_2H_2+2H_2\rightarrow C_2H_6

From the balanced chemical reaction we conclude that,

As, 1 mole of ethane produced from 2 moles of hydrogen

So, 13.78 mole of ethane produced from 13.78\times 2=27.56 moles of hydrogen

Thus, the number of moles of hydrogen needed are, 27.56 mol

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Natali5045456 [20]

Answer:

H2S has a bent geometry, while BeH2 has a linear geometry because: presence of lone pair in H2S

Absence of Lone pair in BeH2

Explanation:

<u>Structure of H2S :</u>

Atomic number of S = 16 = 1s(2 )2s(2) 2p(6) <u>3s(2 )3p(4)</u>

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Valence electron (Involved in bonding) = 6

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<u>Structure of H2S :</u>

Atomic number of Be = 4 = 1s(2 ) 2s(2)

Valence electron (Involved in bonding) = 2

These 2- electron  are involved in bonding with two Hydrogen and form liner shape.*Form sp - hybridized

Hence shape is linear because it has non - lone pair.

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5 0
3 years ago
How many grams of propane are in 20 pounds of propane? Use the conversion 1 lb = 454 g. (Express your answers for the next three
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Answer:

a. =9.1x10^3gC_3H_8

b. 2.1x10^2molC_3H_8

c. Q=-4.6x10^5kJ

Explanation:

Hello,

a. By applying the given information, one obtains:

20lbC_3H_8*\frac{454gC_3H_8}{1lbC_3H_8} =9.1x10^3gC_3H_8

b. By knowing that the propane has a molecular mass of 44g/mol, one obtains:

20lbC_3H_8*\frac{454gC_3H_8}{1lbC_3H_8}*\frac{1molC_3H_8}{44gC_3H_8} =2.1x10^2molC_3H_8

c. Here, the propane's combustion chemical reaction is stated:

C_3H_8+5O_2-->3CO_2+4H_2O

This enthalpy of reaction is computed via:

ΔrH=(3*-393.5kJ/mol)+(4*241kJ/mol)-(-104.7kJ/mol)=-2043kJ/mol

Finally, since it is done for 20 lb of propane (2.1x10^2molC_3H_8), the obtainable energy is:

Q=-2043kJ/mol*2.1x10^2molC_3H_8\\Q=-4.6x10^5kJ

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