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uranmaximum [27]
3 years ago
15

If 498 mol of octane combusts, what volume of carbon dioxide is produced at 23.0 °c and 0.995 atm?

Chemistry
1 answer:
MAXImum [283]3 years ago
5 0
Octane on combustion yields CO₂ and H₂O;

                         2C₈H₁₈<span>  +  2 5O</span>₂    →    16 CO₂<span>  +  18 H</span>₂<span>O
</span>
According to equation,

                      2 moles Octane produces  =  16 moles of CO₂ 
So,
              498 moles Octane will produce  =  X moles of CO₂

Solving for X,
                                     X  =  (498 mol × 16 mol) ÷ 2 mol

                                     X  =  3984 moles of CO₂

Now,
Calculating for Volume,
As,
                                    P V  =  n R T
Solving for V,
                                        V  =  n R T / P       ------- (1)

Pressure  =  P  =  0.995 atm 
<span>Volume  =  V  =  ? </span>
<span>Moles  =  n  =  3984 </span>
<span>Gas Constant  =  R  =  0.0821 L.atm.K</span>⁻¹.<span>mol</span>⁻¹<span> </span>
<span>Temperature  =  T  =  23 + 273 = 296 K
</span>
Putting values in eq. 1

                V  =  (3984 mol × 0.0821 L.atm.mol⁻¹.K⁻¹ × 296 K) ÷ 0.995 atm

                                      Volume  =  973040.95 L
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I think that you are trying to balance this equation.
In order to balance a chemical equation, the numbers of atoms of each element must be equal on both sides of the equation.

In this particular equation, the answer would be (2) HBr + (1) Mg(OH)2 ---> (1) MgBr2 + (2) H2O.

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Explanation:

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       The general equation: AB → A + B.

  • Various methods used in the decomposition of water are -
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Hence, balancing the equation we need to add a coefficient of 2 in front of H_2O on right-hand-side of the equation and  2 in front of H_2 on left-hand-side of the equation.

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