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den301095 [7]
3 years ago
15

What is the percent yield if 107.50 g NH3 reacts with excess O2 according to the

Chemistry
1 answer:
Maksim231197 [3]3 years ago
8 0

Answer:

81.59%

Explanation:

  • 4NH₃ + 5O₂ → 4NO + 6H₂O

First we <u>convert 107.50 g of NH₃ into moles</u>, using its <em>molar mass</em>:

  • 107.50 g NH₃ ÷ 17 g/mol = 6.32 mol NH₃

Now we <u>calculate how many moles of NO would have been formed by the complete reaction of 6.32 moles of NH₃</u>:

  • 6.32 mol NH₃ * \frac{4molNO}{4molNH_3} = 6.32 mol NO

Then we <u>convert 6.32 moles of NO to grams</u>, using its <em>molar mass</em>:

  • 6.32 mol NO * 30 g/mol = 189.60 g NO

Finally we <u>calculate the percent yield</u>:

  • 154.70 g / 189.60 g * 100% = 81.59%

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Let's analyze each reaction.

First, C6H12 has the general formula of an alkene or cycloalkane. However, when we look at the reagents, which are HBr in ROOR, and the final product, we can see that this is an adition reaction where the H and Br were added to a molecule, therefore we can conclude that the initial reactant is an alkene. Now, what happens next? A is reacting with HBr. In general terms when we have an adition of a molecule to a reactant like HBr (Adding electrophyle and nucleophyle) this kind of reactions follows the markonikov's rule that states that the hydrogen will go to the carbon with more hydrogens, and the nucleophyle will go to the carbon with less hydrogen (Atom that can be stabilized with charge). But in this case, we have something else and is the use of the ROOR, this is a peroxide so, instead of follow the markonikov rule, it will do the opposite, the hydrogen to the more substituted carbon and the bromine to the carbon with more hydrogens. This is called the antimarkonikov rule. Picture attached show the possible structure for A. The alkene would have to be the 1-hexene.

Now in the second case we have C, reacting with bromine in light to give also B. C has the formula C6H14 which is the formula for an alkane and once again we are having an adition reaction. In this case, conditions are given to do an adition reaction in an alkane. bromine in presence of light promoves the adition of the bromine to the molecule of alkane. In this case it can go to the carbon with more hydrogen or less hydrogens, but it will prefer the carbon with more hydrogens. In this case would be the terminal hydrogens of the molecules. In this case, it will form product B again. the alkane here would be the hexane. See picture for structures.

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3 years ago
A solution is prepared at that is initially in benzoic acid , a weak acid with , and in sodium benzoate . Calculate the pH of th
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Hello,

In this case, for a concentration of 0.42 M of benzoic acid whose Ka is 6.3x10⁻⁵ in 0.33 M sodium benzoate, we use the Henderson-Hasselbach equation to compute the required pH:

pH=pKa+log(\frac{[base]}{[acid]} )

Whereas the concentration of the base is 0.33 M and the concentration of the acid is 0.42 M, thereby, we obtain:

pH=-log(Ka)+log(\frac{[base]}{[acid]} )\\\\pH=-log(6.3x10^{-5})+log(\frac{0.33M}{0.42M} )\\\\pH=4.1

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