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 ionization energy<span> for </span>hydrogen<span> is 1312 kilojoules per mole. This is the same ... Electrically neutral </span>atoms<span> include a </span>single<span> proton and electron held together.</span>
A general formula of an acid always start with an hydronium ion or H+. The general formula of an acid have to end with an anion. As acids, these compounds are named starting with the prefix "hydro-," then adding the first syllable of the anion, then the suffix "-ic."
Answer:
1.) AgNO₃
2.) 0.563 moles AgBr
Explanation:
The limiting reagent is the reagent that is used up completely during a reaction. It can be identified by calculating which reactant produces the smallest amount of product. This can be done by determining the number of moles of each reagent (via molarity conversion). and then converting it to moles of the product (via mole-to-mole ratio).
AgNO₃ (aq) + KBr (aq) ---> AgBr (s) + KNO₃ (aq)
Molarity (M) = moles / liters
100 mL = 1 L
AgNO₃
45.0 mL / 100 = 45.0 L
1.25 M = ? moles / 0.450 L
? moles = 0.563 moles
KBr
75.0 mL / 100 = 0.750 L
0.800 M = ? moles / 0.750 L
? moles = 0.600 moles
In this case, there is no need to use the mole-to-mole ratio because all of the coefficients are one in the reaction (the amount of the limiting reagent used is the same amount of product produced). Since AgNO₃ produces the smaller amount of product, it is the limiting reagent.
