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Vlad1618 [11]
3 years ago
12

For the following reaction, 5.20 grams of propane (C3H8) are allowed to react with 22.5 grams of oxygen gas. propane (C3H8) (g)

oxygen (g) carbon dioxide (g) water (g) What is the maximum amount of carbon dioxide that can be formed
Chemistry
1 answer:
Sliva [168]3 years ago
4 0

Answer:

15.58g of CO₂ is the maximum amount that can be produced

Explanation:

The propane reacts with oxygen as follows:

C₃H₈ + 5O₂ → 3CO₂ + 4H₂O

<em>Where 1 mole of propane reacts with 5 moles of oxygen</em>

To solve this question we need to find the moles of propane and oxygen to find limiting reactant using the chemical reaction:

<em>Moles propane -Molar mass: 44.1g/mol-:</em>

5.20g * (1mol / 44.1g) = 0.118 moles

<em>Moles oxygen -Molar mass: 32g/mol-:</em>

22.5g O₂ * (1mol / 32g) = 0.703 moles

For a complete reaction of 0.703 moles of oxygen are:

0.703 moles O₂ * (1mol C₃H₈ / 5mol O₂) = 0.141 moles of propane are necessaries. As there are just 0.118 moles of propane, <em>propane is limiting reactant.</em>

The moles of carbon dioxide that are produced are:

0.118 moles C₃H₈ * (3 moles CO₂ / 1 mol C₃H₈) = 0.354 moles CO₂

The maximum mass that can be produced is -Molar mass CO₂: 44.01g/mol-:

0.354 moles CO₂ * (44.01g / mol) =

15.58g of CO₂ is the maximum amount that can be produced

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The standard cell potential Ec for the reduction of silver ions with elemental copper is 0.46V at 25 degrees celsius. calculate
Cloud [144]

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Solution :

The balanced cell reaction will be,  

Cu(s)+2Ag^+(aq)\rightarrow Cu^{2+}(aq)+2Ag(s)

Here, magnesium (Cu) undergoes oxidation by loss of electrons, thus act as anode. silver (Ag) undergoes reduction by gain of electrons and thus act as cathode.

The half oxidation-reduction reaction will be :

Oxidation : Cu\rightarrow Cu^{2+}+2e^-

Reduction : 2Ag^++2e^-\rightarrow 2Ag

Now we have to calculate the Gibbs free energy.

Formula used :

\Delta G^o=-nFE^o

where,

\Delta G^o = Gibbs free energy = ?

n = number of electrons to balance the reaction = 2

F = Faraday constant = 96500 C/mole

E^o = standard e.m.f of cell = 0.46 V

Now put all the given values in this formula, we get the Gibbs free energy.

\Delta G^o=-(2\times 96500\times 0.46)=-88780J/mole

Therefore, the \Delta G for this reaction is, -88780 J/mole.

7 0
3 years ago
There are three isotopes of X element (X):
Nutka1998 [239]

Answer:

Average atomic mass = 17.5 amu.

Explanation:

Given data:

X-17 isotope = atomic mass17.2 amu, abundance:78.99%

X-18isotope =  atomic mass 18.1 amu, abundance 10.00%

X-19isotope = atomic mass:19.1 amu, abundance: 11.01%

Average atomic mass of X = ?

Solution:

Average atomic mass  = (abundance of 1st isotope × its atomic mass) +(abundance of 2nd isotope × its atomic mass) + (abundance of 3rd isotope × its atomic mass)  / 100

Average atomic mass  = (78.99×17.2)+(10.00×18.1) +(11.01+ 19.1) /100

Average atomic mass =  1358.628 + 181 +210.291 / 100

Average atomic mass  = 1749.919 / 100

Average atomic mass = 17.5 amu.

8 0
2 years ago
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iren2701 [21]
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Answer:

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Explanation:

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I hope this helps in any way :)

4 0
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