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UkoKoshka [18]
3 years ago
15

How many grams of Co2 are produced from 12 g of CH4 and 133 g of O2

Chemistry
2 answers:
Andru [333]3 years ago
7 0

Answer:

m_{CO_2}=33gCO_2

Explanation:

Hello,

In this case the undergoing chemical reaction is:

CH_4+2O_2\rightarrow CO_2+2H_2O

Next we identify the limiting reactant with the given amounts of methane and oxygen by computing the available moles of methane and the moles of methane consumed by the 133 g of oxygen:

n_{CH_4}^{available}=12g*\frac{1mol}{16g} =0.75molCH_4\\n_{CH_4}^{consumed\ by\ O_2}=133gO_2*\frac{1molO_2}{32gO_2}*\frac{1molCH_4}{2molO_2}=2.08molCH_4

In such a way, the limiting reactant is methane with 0.75 moles which produce the following mass of carbon dioxide:

m_{CO_2}=0.75molCH_4*\frac{1molCO_2}{1molCH_4} *\frac{44gCO_2}{1molCO_2} \\\\m_{CO_2}=33gCO_2

Best regards.

tankabanditka [31]3 years ago
4 0

Answer:

The correct answer is 33 g CO₂

Explanation:

The reaction between CH₄ and O₂ is a combustion reaction. The complete combustion reaction is described by the following chemical equation:

CH₄(g) + 2 O₂(g)→ CO₂(g) + 2 H₂O(g)

16 g        64 g        44 g           36 g

That means that 1 mol of CH₄ (equal to 16 g) reacts with 2 moles of O₂ (65 g) to give 1 mol of CO₂ (44 g) and 2 moles of H₂O (36 g).

In order to calculate the grams of CO₂ produced by the reaction, we have to figure out which is the <em>limiting reactant</em>. For this, we can choose one reactant and calculate how many grams we need of the other reactant by using the quantity of reactant we have. Let's choose CH₄. According to the chemical equation, 16 g of CH₄ reacts with 64 g of O₂. If we have 12 g, the quantity of O₂ we need is the following:

g O₂ requested = 12 g CH₄ x (64 g O₂)/(16 g CH₄) = 48 g

We need 48 g of O₂ and we have 133 g, so the O₂ is the excess reactant and CH₄ is the limiting reactant. With the limiting reactant, we calculate the amount of product produced. According to the chemical equation, with 16 g of CH₄ we obtain 44 g of CO₂. We have 12 g, so we multiply the grams of CH₄ we have by the factor 44 g CO₂/16 g CH₄:

grams of CO₂ produced = 12 g CH₄ x 44 g CO₂/16 g CH₄ = 33 g CO₂

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See explanation

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Which of the compounds above are strong enough acids to react almost completely with a hydroxide ion (pka of h2o = 15.74) or wit
luda_lava [24]

The compounds can react with OH⁻ and HCO₃⁻ only C₅H₆N pyridinium

<h3><em>Further explanation </em></h3>

In an acid-base reaction, it can be determined whether or not a reaction occurs by knowing the value of pKa or Ka from acid and conjugate acid (acid from the reaction)

Acids and bases according to Bronsted-Lowry

Acid = donor (donor) proton (H + ion)

Base = proton (receiver) acceptor (H + ion)

If the acid gives (H +), then the remaining acid is a conjugate base because it accepts protons. Conversely, if a base receives (H +), then the base formed can release protons and is called the conjugate acid from the original base.

From this, it can be seen whether the acid in the product can give its proton to a base (or acid which has a lower Ka value) so that the reaction can go to the right to produce the product.

The step that needs to be done is to know the pKa value of the two acids (one on the left side and one on the right side of the arrow), then just determine the value of the equilibrium constant

Can be formulated:

K acid-base reaction = Ka acid on the left : K acid on the right.

or:

pK = acid pKa on the left - pKa acid on the right

K = equilibrium constant for acid-base reactions

pK = -log K;

K~=~10^{-pK}

K value> 1 indicates the reaction can take place, or the position of equilibrium to the right.

There is some data that we need to complete from the problem above, which is the pKa value of some compounds that will react, namely:

pyridinium pKa = 5.25

acetone pKa = 19.3

butan-2-one pKa = 19

Let's look at the K value of each possible reaction:

pka H₂O = 15.74, pka of H₂CO₃ = 6.37)

  • 1. C₅H₆N pyridinium

* with OH⁻

C₅H₆N + OH- ---> C₅H₅N- + H₂O

pK = pKa pyridinium - pKa H₂O

pK = 5.25 - 15.74

pK = -10.49

K~=~10^{4.9}

K values> 1 indicate the reaction can take place

* with HCO3⁻

C₅H₆N + HCO₃⁻-- ---> C₅H₅N⁻ + H₂CO₃

pK = 5.25 - 6.37

pK = -1.12

K`=~10^{1.12]

Reaction can take place

  • 2. Acetone C₃H₆O

* with OH-

C₃H₆O + OH⁻ ---> C₃H₅O- + H₂O

pK = 19.3 - 15.74

pK = 3.56

K~=~10^{ -3.56}

Reaction does not happen

* with HCO₃-

C₃H₆O + HCO₃⁻ ----> C₃H₅O⁻ + H₂CO₃

pK = 19.3 - 6.37

pK = 12.93

K`=~10 ^{-12.93}

Reaction does not happen

  • 3. butan-2-one C₄H₇O

* with OH-

C₄H₇O + OH- ---> C₄H₆O- + H₂O

pK = 19 - 15.74

pK = 3.26

K~=~10^{-3.26}

Reaction does not happen

* with HCO₃⁻

C₄H₇O + HCO₃⁻ ---> C₄H₆O⁻ + H₂CO₃

pK = 19 - 6.37

pK = 12.63

K~=~ 10^{-12.63}

Reaction does not happen

So that can react with OH⁻ and HCO₃⁻ only C₅H₆N pyridinium

<h3><em>Learn more </em></h3>

the lowest ph

brainly.com/question/9875355

the concentrations at equilibrium.

brainly.com/question/8918040

the ph of a solution

brainly.com/question/9560687

Keywords : acid base reaction, the equilibrium constant

5 0
3 years ago
Read 2 more answers
why is a molecule of CO2 non polar even though the bonds between the carbon atom and the oxygen atoms are polar
stira [4]

Answer:

m

Explanation:

m

5 0
3 years ago
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