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elena-s [515]
2 years ago
15

How many grams of CO2 will be produced when 8.50 g of methane react with 15.9 g of O2, according to the following reaction? 
CH4

(g) +2O2(g)→CO2(g) +2H2O(g) 



Chemistry
1 answer:
Aleonysh [2.5K]2 years ago
8 0

The amount of CO_2 that would be produced will be 10.93 g

<h3>Stoichiometric calculations</h3>

From the equation of the reaction, the mole ratio of methane to oxygen is 1:2.

Mole of 8.50 g methane = 8.50/16.04 = 0.53 moles

Mole of 15.9 g oxygen = 15.9/32 = 0.4969 moles

Thus, methane is in excess while oxygen is limiting.

Mole ratio of O_2 and  CO_2  = 2:1

Equivalent mole of  CO_2  = 0.4969/2 = 0.25 moles

Mass of 0.25 moles  CO_2  = 0.25 x 44.01 = 10.93 g

More on stoichiometric calculations can be found here: brainly.com/question/27287858

#SPJ1

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A hypothetical element has an atomic weight of 48.68 amu. It consists of three isotopes having masses of 47.00 amu, 48.00 amu, a
Morgarella [4.7K]

Answer : The percent abundance of the heaviest isotope is, 78 %

Explanation :

Average atomic mass of an element is defined as the sum of masses of each isotope each multiplied by their natural fractional abundance.

Formula used to calculate average atomic mass follows:

\text{Average atomic mass }=\sum_{i=1}^n\text{(Atomic mass of an isotopes)}_i\times \text{(Fractional abundance})_i

As we are given that,

Average atomic mass = 48.68 amu

Mass of heaviest-weight isotope = 49.00 amu

Let the percentage abundance of heaviest-weight isotope = x %

Fractional abundance of heaviest-weight isotope = \frac{x}{100}

Mass of lightest-weight isotope = 47.00 amu

Percentage abundance of lightest-weight isotope = 10 %

Fractional abundance of lightest-weight isotope = \frac{10}{100}

Mass of middle-weight isotope = 48.00 amu

Percentage abundance of middle-weight isotope = [100 - (x + 10)] %  = (90 - x) %

Fractional abundance of middle-weight isotope = \frac{(90-x)}{100}

Now put all the given values in above formula, we get:

48.68=[(47.0\times \frac{10}{100})+(48.0\times \frac{(90-x)}{100})+(49.0\times \frac{x}{100})]

x=78\%

Therefore, the percent abundance of the heaviest isotope is, 78 %

5 0
3 years ago
Read 2 more answers
HELP! ASAP!
harina [27]

Given the model from the question,

  • The products are: N₂, H₂O and H₂
  • The reactants are: H₂ and NO
  • The limiting reactant is H₂
  • The balanced equation is: 3H₂ + 2NO —> N₂ + 2H₂O + H₂

<h3>Balanced equation </h3>

From the model given, we obtained the ffolowing

  • Red => Oxygen
  • Blue => Nitrogen
  • White => Hydrogen

Thus, we can write the balanced equation as follow:

3H₂ + 2NO —> N₂ + 2H₂O + H₂

From the balanced equation above,

  • Reactants: H₂ and NO
  • Product: N₂, H₂O and H₂

<h3>How to determine the limiting reactant</h3>

3H₂ + 2NO —> N₂ + 2H₂O + H₂

From the balanced equation above,

3 moles of H₂ reacted with 2 moles of NO.

Therefore,

5 moles of H₂ will react with = (5 × 2) / 3 = 3.33 moles of NO

From the calculation made above, we can see that only 3.33 moles of NO out of 4 moles given are required to react completely with 5 moles of H₂.

Thus, H₂ is the limiting reactant

Learn more about stoichiometry:

brainly.com/question/14735801

#SPJ1

8 0
2 years ago
Balance the following and label what type of reaction is taking place:
Vadim26 [7]

4,27,20,18

I hope this helps.

3 0
3 years ago
What type of mixture is this salad dressing
melisa1 [442]
I believe that would be oil and vinegar.

5 0
3 years ago
Read 2 more answers
What type of reaction is "Separating Water"?
Lostsunrise [7]

Answer:

Decomposition

3 0
3 years ago
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