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faltersainse [42]
3 years ago
7

How many moles of carbon dioxide gas can be produced when 12.8 moles of benzene (C6H6) react with excess oxygen? Unbalanced equa

tion: C6H6 + O2 → CO2 + H2O Show, or explain, all of your work along with the final answer.
Chemistry
2 answers:
swat323 years ago
6 0

Answer: 76.8 mole

Explanation:

  • Firstly, the balanced equation of reacting benzene with excess oxygen is:

2C6H6 + 15O2 → 6H2O + 12CO2

  • It is clear that every two moles of benzene react with 15 mole of oxygen to produce 6 moles of water and 12 mole of CO2
  • Which means that 2 moles of benzene give 12 moles of CO2
  • So, n(C6H6)/2 = n(CO2)/12
  • which give n(C6H6) = n(CO2)/6
  • n(CO2) = n(C6H6) x 6 = 12.8 x 6 = 76.8 mole
joja [24]3 years ago
6 0

2C6H6 + 15O2 → 12CO2 + 6H2O   is the balanced equation.

So 2 moles of benzene produces 12 moles of CO2.

Therefore 1 Mole of benzene produces  6 moles of CO2,

and 12.8 moles of benzene produces 12.8 * 6 = 76.8 moles of CO2, (answer).



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Therefore, 1 liter is 1/224 moles

one mole of nitrogen 14 is 14

Therefore 1 liter of the nitrogen weighs 1/224×14

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Calculate the number of moles in a 14.5 gram sample of C4H10.
cricket20 [7]
Moles= mass divided by molar mass
Molar mass= 12.01(4) + 1.01(10)
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Moles=14.5g / 58.14g/mol
         =0.249

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6 0
2 years ago
If 1.02 g of nickel reacted with 750. mL of 0.112 M hydrobromic acid, how much of each will be present at the end of the reactio
kati45 [8]

Answer:

35.1% is percent yield

Explanation:

<em>Full question: Assume no volume change.  If you formed 0.0910 atm of gas, what is the percent yield?</em>

<em />

The reaction that is occurring is:

Ni + 3HBr → NiBr₃ + 3/2H₂(g)

First, we will determine moles of Ni and HBr to determine limiting reactant and theoretical yield

Using ideal gas law, we can determine the moles of hydrogen formed. Thus, we can find percent yield:

<em>Moles Ni (Molar mass: 58.69g/mol):</em>

1.02g * (1mol / 58.69g) = 0.01738moles Ni

<em>Moles HBr:</em>

0.750L * (0.112mol/L) = 0.084 moles of HBr.

For a complete reaction of the 0.084 moles of HBr you need:

0.084mol HBr * (1 mole Ni / 3 moles HBr) = 0.028 moles of Ni.

As there are just 0.01738 moles of Ni, the Ni is limiting reactant. Assuming a theoretical yield, moles of H₂ produced are:

0.01738moles Ni * (3/2 H₂ / 1 mol Ni) = 0.02607 moles H₂

Now, moles of H₂ produced are:

PV = nRT

PV/RT = n

<em>Where P is pressure (0.0910atm)</em>

<em>V is volume (2.50L)</em>

<em>R is gas constant (0.082atmL/molK)</em>

<em>T is absolute temperature in Kelvin (30°C + 273.15 = 303.15K)</em>

<em>And n are moles</em>

PV/RT = n

0.0910atm*2.50L/0.082atmL/molK*303.15K = n

0.00915 moles = n

<em />

And percent yield (Produced moles / Theoretical moles * 100) is:

0.00915 moles / 0.02607moles =

<h3>35.1% is percent yield</h3>
8 0
2 years ago
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