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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 and an excess of nitrogen? Be sure to balance the equation before you begin and round to the proper accuracy. Please leave any units off.

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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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