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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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8 0
2 years ago
1.How many mL of 0.401 M HI are needed to dissolve 5.97 g of BaCO3?
garri49 [273]

Answer:

The answer to your question is:

1.- volume = 0.151 l or 151 ml

2.- 0.241 l  or 241 ml of NaOH

Explanation:

1.-

Data

V = ? HI = 0.401 M

BaCO3 = 5.97 g

                     2HI(aq)    +    BaCO3(s)   ⇒   BaI2(aq) + H2O(l) + CO2(g)

MW BaCO3 = 137 + 12 + 48 = 197 g

                     197 g of BaCO3 ----------------- 1 mol

                     5.97 g                -----------------   x

                     x = (5.97 x 1) /197

                    x = 0.03 mol of BaCO3

                    2 moles of HI ----------------  1 mol of BaCO3

                    x                     ----------------  0.03 mol of BaCO3

                    x = (0.03 x 2) / 1

                   x = 0.060 mol of HI

Molarity = moles / volume

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volume = 0.060 / 0.401

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2.-

V = ?    NaoH 0.757 M

Co⁺² Volume = 167 ml   0.548 M

             CoSO4(aq) + 2NaOH(aq)   ⇒   Co(OH)2(s) + Na2SO4(aq)

Moles of Co = Molarity x  volume

Moles of Co = 0.548 x 0.167

Moles of Co = 0.092

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                                0.092 moles      ---------------   x

                                x = (0.092 x 2) /1

                               x = 0.183 moles of NaOH

Volume of NaOH = moles / molarity

                             = 0.183 / 0.757

                            = 0.241 l  or 241 ml of NaOH

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