The total percent yield:
After the combustion reaction with methane, the percent yield was 66.7%.
Combustion of Methane:
- Methane produces a blue flame as it burns in the atmosphere.
- Methane burns in the presence of enough oxygen to produce carbon dioxide (CO₂) and water (H₂O).
- It creates a significant quantity of heat during combustion, making it an excellent fuel source.
The other reactant, air's excess oxygen, is always present, making methane the limiting reactant. As a result, the amount of CH₄ burned will determine how much CO₂ and H₂O are produced.
The following chemical process produces carbon dioxide from methane:
CH₄ + 2O₂ ⇒ CO₂ + 2H₂O
Calculations:
1. <u><em>Theoretical quantity of carbon dioxide:</em></u>
All calculations will be based on the amount of methane because the problem specifies that it is the limiting reagent:
12.0g of CH₄ × (1 mol of CH₄/16g CH₄) × (1 mole of CO₂/1 mole of CH₄) × (44g CO₂/1 mole of CO₂)
= 33g of CO₂
2. <u><em>Percent yield:</em></u>
= Actual yield/Theoretical yield × 100
= 22.0g/33g × 100
= 66.7%
Learn more about the percent yield here,
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The molecular formula is D. C_8H_20O_4Si.
<em>Step 1</em>.Calculate the <em>empirical formula
</em>
a) Calculate the moles of each element
Moles of C= 196.01 g C × (1 mol C/12.01 g C) = 16.325 mol C
Moles of H = 41.14 g H × (1 mol H/1.008 g H) = 40.813 mol H
Moles of O = 130.56 g O × (1 mol O/16.00 g O) = 8.1650 mol O
Moles of Si = 57.29 g Si × (1 mol Si/28.085 g Si) = 2.0399 mol Si
b) Calculate the molar ratio of each element
Divide each number by the smallest number of moles and round off to an integer
C:H:O:Si = 8.0027:20.008:4.0027:1 ≈ 8:20:4:1
c) Write the empirical formula
EF = C_8H_20O_4Si
<em>Step </em>2. Calculate the <em>molecular formula</em>
EF Mass = 208.33 u
MF mass = 208.329 u
MF = (EF)_n
n = MF Mass/EF Mass = 208.329 u/208.33 u = 1.0000 ≈ 1
MF = C_8H_20O_4Si
3,2mol * 2 = 6,4mol of C atoms
Answer:
oxidation- reduction
Explanation:
where gaining electronic reduces one element and losing them oxidize the other nitric acid is not only strong it is also a oxidizing agent
<h2>Oxidize: copper = Cu+2</h2>
55.9 kPa; Variables given = volume (V), moles (n), temperature (T)
We must calculate <em>p</em> from <em>V, n</em>, and <em>T</em>, so we use <em>the Ideal Gas Law</em>:
<em>pV = nRT</em>
Solve for <em>p</em>: <em>p = nRT/V</em>
R = 8.314 kPa.L.K^(-1).mol^(-1)
<em>T</em> = (265 + 273.15) K = 538.15 K
<em>V</em> = 500.0 mL = 0.5000 L
∴ <em>p</em> = [6.25 x 10^(-3) mol x 8.314 kPa·L·K^(-1)·mol^(-1) x 538.15 K]/(0.5000 L) = 55.9 kPa
