For Iron: 
For Oxygen:
These are the two chemical symbols for the two elements found in Iron Oxide.
For Oxygen:
These are the two chemical symbols for the two elements found in Iron Oxide.
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1 answer:
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<u>Answer:</u> The theoretical yield of acetanilide is 6.5 grams.
<u>Explanation:</u>
To calculate the number of moles, we use the equation:
.....(1)
- <u>For aniline:</u>
Given mass of aniline = (We know that:
)
Molar mass of aniline = 93.13 g/mol
Putting values in equation 1, we get:
- <u>For acetic anhydride:</u>
To calculate the mass of acetic anhydride, we use the equation:
Volume of acetic anhydride =
Density of acetic anhydride = 1.08 g/mL
Putting values in above equation:
Given mass of acetic anhydride = 6.08 g
Molar mass of acetic anhydride = 102.1 g/mol
Putting values in equation 1, we get:
The chemical equation for the reaction of aniline and acetic anhydride follows:
By Stoichiometry of the reaction:
1 mole of aniline reacts with 1 mole of acetic anhydride
So, 0.048 moles of aniline will react with = of acetic anhydride
As, given amount of acetic anhydride is more than the required amount. So, it is considered as an excess reagent.
Thus, aniline is considered as a limiting reagent because it limits the formation of product.
By Stoichiometry of the reaction:
1 mole of aniline produces 1 mole of acetanilide
So, 0.048 moles of aniline will produce = of acetanilide
Now, calculating the theoretical yield of acetanilide by using equation 1:
Moles of acetanilide = 0.048 moles
Molar mass of acetanilide = 135.17 g/mol
Putting values in equation 1, we get:
Hence, the theoretical yield of acetanilide is 6.5 grams.
Answer:
Reagent O₂ will be consumed first.
Explanation:
The balanced reaction between O₂ and C₄H₁₀ is:
2 C₄H₁₀ + 13 O₂ → 8 CO₂ + 10 H₂O
Then, by reaction stoichiometry, the following amounts of reactants and products participate in the reaction:
- C₄H₁₀: 2 moles
- O₂: 13 moles
- CO₂: 8 moles
- H₂O: 10 moles
Being:
- C: 12 g/mole
- H: 1 g/mole
- O: 16 g/mole
The molar mass of the compounds that participate in the reaction is:
- C₄H₁₀: 4*12 g/mole + 10*1 g/mole= 58 g/mole
- O₂: 2*16 g/mole= 32 g/mole
- CO₂: 12 g/mole + 2*16 g/mole= 44 g/mole
- H₂O: 2*1 g/mole + 16 g/mole= 18 g/mole
Then, by reaction stoichiometry, the following mass quantities of reactants and products participate in the reaction:
- C₄H₁₀: 2 moles* 58 g/mole= 116 g
- O₂: 13 moles* 32 g/mole= 416 g
- CO₂: 8 moles* 44 g/mole= 352 g
- H₂O: 10 moles* 18 g/mole= 180 g
If 78.1 g of O₂ react, it is possible to apply the following rule of three: if by stoichiometry 416 g of O₂ react with 116 g of C₄H₁₀, 62.4 g of C₄H₁₀ with how much mass of O₂ do they react?
mass of O₂= 223.78 grams
But 21.78 grams of O₂ are not available, 78.1 grams are available. Since you have less mass than you need to react with 62.4 g of C₄H₁₀, <u><em>reagent O₂ will be consumed first.</em></u>