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pav-90 [236]
3 years ago
13

Plz help guys ASAP! Thanks in advance

Chemistry
1 answer:
MariettaO [177]3 years ago
4 0

Answer:

3.6 moles

Explanation:

The balanced equation for the reaction is given below:

N₂ + 3H₂ —> 2NH₃

From the balanced equation above,

1 mole of N₂ reacted with 3 moles of H₂ to produce 2 moles NH₃.

Next, we shall determine the limiting reactant. This can be obtained as follow:

From the balanced equation above,

1 mole of N₂ reacted with 3 moles of H₂.

Therefore, 3.2 moles of N₂ will react with = 3.2 × 3 = 9.6 moles of H₂.

From the calculation made above, we can see that it will take a higher amount (i.e 9.6 moles) of H₂ than what was given (i.e 5.4 moles) to react completely with 3.2 moles of N₂.

Therefore, H₂ is the limiting reactant and N₂ is the excess reactant.

Finally, we the greatest quantity of ammonia, NH₃ produced from the reaction.

In this case, the limiting reactant will be use because all of it is consumed in the reaction.

The limiting reactant is H₂ and greatest quantity of ammonia, NH₃ produced can be obtained as follow:

From the balanced equation above,

3 moles of H₂ reacted to produce 2 moles NH₃.

Therefore, 5.4 of H₂ will react to produce = (5.4 × 2)/3 = 3.6 moles of NH₃

Thus, the greatest quantity of ammonia, NH₃ produced from the reaction is 3.6 moles

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Answer:

SPECIFIC HEAT CAPACITY

Explanation:

Q = MC DELTA T

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c = specific heat

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4 0
3 years ago
How many dm3 of oxygen at STP would be required to react completely with 38.8g of propane? (3 marks)
11Alexandr11 [23.1K]

98.56 dm^3 of oxygen at STP would be required to react completely with 38.8g of propane.

<u>Given that :</u>

molar mass of propane = 44 g/mol

mass of propane = 38.8 g

∴ Moles present in 38.8 g of propane = \frac{38.8}{44} = 0.88 mole

<u>applying rule of balanced equations </u>

1 mole of propane = 5 moles of oxygen

0.88 mole of propane =  5 * 0.88 = 4.4 moles of oxygen

Note : volume of 1 mole of oxygen at STP = 22.4 dm^3

∴Total volume of oxygen required at STP = 22.4 * 4.4 = 98.56 dm^3

Hence we can conclude that the volume of oxygen at STP required to react completely 98.56 dm^3

Learn more : brainly.com/question/16998374

5 0
1 year ago
Aluminum and oxygen react according to the following equation: 4Al + 3O2 -&gt; 2Al2O3 In a certain experiment, 4.6g Al was react
stiv31 [10]

Answer:

Percent yield: 78.2%

Explanation:

Based on the reaction:

4Al + 3O₂ → 2Al₂O₃

<em>4 moles of Al produce 2 moles of Al₂O₃</em>

<em />

To find percent yield we need to find theoretical yield (Assuming a yield of 100%) and using:

(Actual yield (6.8g) / Theoretical yield) × 100

Moles of 4.6g of Al (Molar mass: 26.98g/mol) are:

4.6g Al × (1mol / 26.98g) = 0.1705 moles of Al.

As 4 moles of Al produce 2 moles of Al₂O₃, theoretical moles of Al₂O₃ obtained from 0.1705 moles of Al are:

0.17505 moles Al × (2 moles Al₂O₃ / 4 moles Al) = <em>0.0852 moles of Al₂O₃</em>,

In grams (Molar mass Al₂O₃ = 101.96g/mol):

0.0852 moles of Al₂O₃ × (101.96g / mol) =

<h3>8.7g of Al₂O₃ can be produced (Theoretical yield)</h3>

Thus, Percent yield is:

(6.8g / 8.7g) × 100 =

<h3>78.2% </h3>
8 0
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iren2701 [21]

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Classification will be Potassium, Bromine, and Argon

Explanation:

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?

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