What is the percent yield if 107.50 g NH3 reacts with excess O2 according to the
1 answer:
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 *
= 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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Answer:
Explanation:
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A and C reacts with two differents reagents and conditions, however both of them gives the same product.
Let's analyze each reaction.
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Answer:
Explanation:
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:
Whereas the concentration of the base is 0.33 M and the concentration of the acid is 0.42 M, thereby, we obtain:
Regards.