CHEMISTRY HIGH SCHOOL

One way to remove nitrogen oxide from smokestack emissions is to react it with ammonia: 4 NH3 (g) + 6 NO (g) ----------> 5 N2 (g) + 6H2O (I)Fill in the blanks below:

12.3 mol of NO reacts with ______ mol of ammonia
5.87 mol of NO yields _______ mol nitrogen

Answers

Answer 1
Answer:

Final answer:

12.3 moles of nitrogen oxide (NO) react with 8.2 moles of ammonia (NH3) in the given reaction. The reaction of 5.87 moles of NO yields 4.89 moles of nitrogen (N2).

Explanation:

The overall reaction equation represents an ideal stoichiometric ratio where 4 moles of ammonia (NH3) react with 6 moles of nitrogen oxide (NO) to produce 5 moles of nitrogen (N2) and 6 moles of water (H2O). Let's work with this ratio to fill in the blanks of your question.

Given that 6 moles of NO react with 4 moles of NH3, we can set up a ratio to find the unknown quantity of NH3 that reacts with 12.3 moles of NO:  (4 NH3 / 6 NO) x 12.3 NO = 8.2 moles of NH3

Similarly, 5 moles of nitrogen are produced per 6 moles of NO reacted, so we can find out how much nitrogen is produced from 5.87 moles of NO: (5 N2 / 6 NO) x 5.87 NO = 4.89 moles of nitrogen.

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Answer 2
Answer:

Final answer:

12.3 moles of NO reacts with 8.2 moles of Ammonia. 5.87 moles of NO yields 4.89 moles of Nitrogen.

Explanation:

In these types of problems, we use the concept of mole ratios from the balanced chemical equation. According to the reaction, 4 moles of NH3 react with 6 moles of NO to yield 5 moles of N2. Therefore, for 12.3 moles of NO, we apply the ratio 4/6 to determine that it reacts with 8.2 moles of NH3.

Similarly, 5.87 moles of NO, based on the ratio 5/6, yields 4.89 moles of N2. So, these are the missing quantities for the reaction.

Learn more about Mole Ratios here:

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what volume of a 0.716 m kbr solution is needed to provide 25.5 g of kbr? 35.7 ml 14.7 ml 299 ml 0.299 ml 153 ml

Answers

Answer:

\huge{ \boxed{299 \: ml}}

Explanation:

The volume of KBr required can be found by using the formula;

v = (m)/(M * c)

c is the concentration in M , mol/dm³ or mol/L

v is the volume in L or dm³

m is the mass in grams

M is the molar mass in g/mol

From the question;

m = 25.5 g

c = 0.716

Molar mass (M) of KBr = 39 + 79.9 = 118.9 g/mol

\therefore \: v = (25.5)/(118.9 * 0.716) \n = 0.299 \: l \n \n \text{but} \: 1 \: l \: = 1000 \: ml \n \therefore0.299l \: = (0.299)/(1) * 1000 \: ml \n = 299 \: ml \n \therefore \: v = 299 \: ml

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Answers

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