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Tamiku [17]
3 years ago
8

The equilibrium constant for the reaction of carbon monoxide with water is 1.845. if 1.00 mol of each reactant is placed in a 2.

00 l container, what is the concentration of carbon dioxide at equilibrium? co(g) + h2o(g) co2(g) + h2(g)
Chemistry
1 answer:
SpyIntel [72]3 years ago
4 0
[CO] = 1 mol / 2L = 0.5 M

[
According to the equation:

and by using the ICE table:

             CO(g) + H2O(g) ↔   CO2(g) + H2(g)

initial     0.5            0.5                    0          0

change  -X              -X                   +X         +X
     
Equ       (0.5-X)       (0.5-X)                     X            X

when Kc = X^2 * (0.5-X)^2

by substitution:

1.845 = X^2 * (0.5-X)^2  by solving for X 

∴X = 0.26

∴ [CO2] = X = 0.26
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The maximum mass of B₄C that can be formed from 2.00 moles of boron (III) oxide is 55.25 grams.

<h3>What is the stoichiometry?</h3>

Stoichiometry of the reaction gives idea about the relative amount of moles of reactants and products present in the given chemical reaction.

Given chemical reaction is:

2B₂O₃ + 7C → B₄C + 6CO

From the stoichiometry of the reaction, it is clear that:

2 moles of B₂O₃ = produces 1 mole of B₄C

Now mass of B₄C will be calculated by using the below equation:

W = (n)(M), where

  • n = moles = 1 mole
  • M = molar mass = 55.25 g/mole

W = (1)(55.25) = 55.25 g

Hence required mass of B₄C is 55.25 grams.

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Calculate the mass of Octane needed to release 6.20 mol Co2
n200080 [17]
The combustion reaction of octane is as follow,

                           C₈H₁₈  +  25/2 O₂     →     8 CO₂  +  9 H₂O

According to balance equation,

8 moles of CO₂ are released when  =  114.23 g (1 mole) Octane is reacted

So,

      6.20 moles of CO₂ will release when  =  X g of Octane is reacted

Solving for X,
                                     X  =  (114.23 g × 6.20 mol) ÷ 8 mol

                                     X  =  88.52 g of Octane
Result:
           88.52 g of Octane is needed to release 6.20 mol CO₂.
8 0
3 years ago
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