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kari74 [83]
3 years ago
5

If the reaction below takes place at STP how many L of ammonia (NH3) will be formed if one starts with 3.45 moles of hydrogen an

d an excess of nitrogen? Be sure to balance the equation before you begin and round to the proper accuracy. Please leave any units off.
Chemistry
1 answer:
lesya692 [45]3 years ago
6 0

Answer:

51.5 L of ammonia

Explanation:

Ammonia is produced according to this equation:

N₂(g) + 3H₂(g) → 2NH₃ (g)

Let's propose this rule of three:

3 moles of hydrogen can produce 2 moles of ammonia

Then 3.45 moles of hydrogen, will produce (3.45 . 2) / 3 = 2.3 moles of NH₃

We apply the Ideal Gases Law at STP to determine the volume

STP = 1 atm and 273K

1 atm . V = 2.3 mol . 0.082 . 273K

V = (2.3 mol . 0.082 . 273K) / 1 atm = 51.5 L

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3 years ago
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makvit [3.9K]

Answer:

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Best known systems:

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Katena32 [7]

Answer:

36.4 atm

Explanation:

To find the pressure, you need to use the Ideal Gas Law. The equation looks like this:

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In this equation,

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-----> n = moles

-----> R = constant (0.0821 L*atm/mol*K)

-----> T = temperature (K)

Before you can plug the given values into the equation, you first need to convert Celsius to Kelvin.

P = ? atm                         R = 0.0821 L*atm/mol*K

V = 5.00 L                       T = 393 °C + 273.15 = 312.45 K

n = 7.10 moles

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P(5.00 L) = (7.10 moles)(0.0821 L*atm/mol*K)(312.45 K)

P(5.00 L) = 182.130

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true

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