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julsineya [31]
3 years ago
7

If you feed 100 kg of N2 gas and 100 kg of H2 gas into a reactor. What is the limiting reactant?

Chemistry
1 answer:
mixas84 [53]3 years ago
8 0

Answer:

N₂ is the limiting reactant

Explanation:

The balanced reaction between N₂ gas and H₂ gas is:

N₂ + 3H₂ → 2NH₃

In order to determine the limiting reactant, we have to <u>calculate the number of moles of each rectant</u>, using their molecular weight:

  • Moles of N₂= 100 kg * \frac{1kmol}{28kg} = 3.57 kmol
  • Moles of H₂= 100 kg * \frac{1kmol}{2kg} = 50.0 kmol

Lastly, we multiply the number of moles of N₂ by 3, and the number of moles of H₂ by 1; due to the coefficients in the balanced reaction. Whichever number is lower, belongs to the limiting reactant.

N₂ => 10.7

H₂ => 50.0

Thus N₂ is the limiting reactant

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You have a radioactive isotope with a mass of 400 grams. The sample has a half life of 5 days. How much is left after 25 days ha
Alexeev081 [22]

Answer:

Amount left after 25 days = 12.5 g

Explanation:

Given data:

Mass of sample = 400 g

Half life of sample = 5 days

Mass left after 25 days = ?

Solution:

First of all we will calculate the number of half lives passes in given time period.

Number of half lives = Time elapsed / Half life

Number of half lives = 25 days/ 5 days

Number of half lives = 5

At time zero = 400 g

At 1st half life = 400 g/2 = 200 g

At 2nd half life = 200 g/2 = 100 g

At 3rd half life = 100 g/2 = 50 g

At 4th half life = 50 g/2 = 25 g

At 5th half life = 25 g/2 = 12.5 g

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3 years ago
What is chemical nitrogen​
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4 0
3 years ago
C) Do grits work better at killing fire ants than commercial products?
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6 0
3 years ago
Give the percent yield when 28.16 g of CO2 are formed from the reaction of 8.000 moles of C8H18 with 16.00 moles of O2.
kipiarov [429]

Answer:

Percent yield of CO₂ is 6.25 %.

Explanation:

Given data:

Percent yield of CO₂ = ?

Actual yield of CO₂ = 28.16 g

Number of moles of C₈H₁₈ = 8.000 mol

Number of moles of O₂ = 16.00 mol

Solution:

Chemical equation:

2C₈H₁₈ + 25O₂  →  16CO₂ + 18H₂O

Now we will compare the moles of C₈H₁₈ and O₂  with CO₂.

                 C₈H₁₈            :          CO₂

                    2                :             16

                  8.000          :            16/2×8.000 = 64 mol

                   O₂               :             CO₂

                    25              :              16

                     16              :              16/25×16= 10.24 mol

Less number of moles  of CO₂ are produced from 16 moles of O₂. it will limit the yield of CO₂.

Grams of CO₂ produced:

Mass = number of moles × molar mass

Mass = 10.24 mol  × 44 g/mol

Mass = 450.56 g

Percentage yield of CO₂:

Percentage yield = actual yield / theoretical yield × 100

Percentage yield = 28.16 g/ 450.56 g× 100

Percentage yield = 6.25 %

               

   

7 0
3 years ago
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