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Lyrx [107]
3 years ago
11

Consider the reaction: CH4 + 2O2 = CO2 + 2H2O

Chemistry
1 answer:
crimeas [40]3 years ago
4 0

Answer:

The answer to your question is:

a)  80 g of O2

b) O2, 15.13 g of CO2

c) It's not posible to know which is the limiting reactant.

Explanation:

Reaction                             CH4   +   2O2   ⇒   CO2   +   2H2O

a. Calculate the grams of O2 needed to react with 20.00 grams of CH4. _____________

MW CH4 = 16 g

MW O2 = 32 g

                               16 g of CH4 ----------------  2(32) g of O2

                               20 g              --------------    x

                               x = (20 x 64) / 16 = 80 g of O2

b. Given 15.00 g. of CH4 and 22.00 g. of O2, identify the limiting reactant and calculate the grams of CO2 that can be produced. LR _________ grams CO2 _________ .  

                                CH4   +   2O2   ⇒   CO2   +   2H2O

                                15 g         22 g

                                16 g of CH4 ----------------  64 g of O2

                                15 g of CH4  ---------------   x

                               x = (15 x 64) / 16 = 60 g of O2

The Limiting reactant is O2 because it is necessary 60g of O2 for 16 g of CH4 and there are only 22.

                                 CH4   +   2O2   ⇒   CO2   +   2H2O

                        64 g of O2 ------------------  44 g of CO2

                        22 g of O2 ------------------   x

                        x = (22 x 44)/ 64 = 15. 13 g of CO2

c. For the reaction CH4 + 2O2 = CO2 + 2H2O, if you have 10.31 g. of CH4 and an unknown amount of oxygen, and form 20.00 g. of CO2, i. Identify if there is a limiting reactant ______________ ii. Calculate the number of grams of the limiting reactant present if there is one. ______________  

                           CH4   +   2O2   ⇒   CO2   +   2H2O                              

                           10.31 g                     20 g

We can identify the limiting reactant if we know the quantity of the reactants, if we only know the quantity of one it is not posible to which is the limiting reactant.

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83.6 kg KClO3 (1000g / 1kg) </span>(1 mol / 122.50g )( 3 mol O2 / 2 mol KClO3 ) ( 32 g O2 / 1 mol O2 ) = 3275.76 g O2
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The amount, in mL, of the concentrated acid required, would be 1.1875 mL

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From the dilution equation:

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m2 = 0.285M, m1 = 12.0M v2 = 50.0 mL

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Teniendo en cuenta la ley de Boyle, el volumen del oxigeno a 415 mmHg, si la temperatura permanece constante, es 7,69 L.

Al aumentar el volumen, las partículas (átomos o moléculas) del gas tardan más en llegar a las paredes del recipiente y por lo tanto chocan menos veces por unidad de tiempo contra ellas. Esto significa que la presión será menor porque ésta representa la frecuencia de choques del gas contra las paredes.

De esta manera se relaciona la presión y el volumen, determinando la ley de Boyle que dice:

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La ley de Boyle se expresa matemáticamente como:

P×V=k

Teniendo un estado inicial 1 y final 2, se cumple:

P₁×V₁= P₂×V₂

En este caso, se sabe que:

  • P₁= 760 mmHg
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  • P₂= 415 mmHg
  • V₂= ?

Reemplazando en la expresión matemática para la Ley de Boyle:

760 mmHg× 4.2 L= 415 mmHg× V₂

Resolviendo:

(760 mmHg× 4.2 L)÷ 415 mmHg= V₂

<u><em>7,69 L= V₂</em></u>

Finalmente, el volumen del oxigeno a 415 mmHg, si la temperatura permanece constante, es 7,69 L.

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