How many grams of oxygen are needed to react completely with 200.0 g of ammonia, nh3? 4nh3(g) + 5o2(g) ®4no(g) + 6h2o(g)
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Answer:
12 atm
Explanation:
First, let us convert Celcius into Kelvin: 28.0 °C = 301.15 K and 129.0 °C = 402.15 K
For this question we must employ the Combined Gas Law: , where
is the initial pressure and
is the new pressure.
We know that intitially, P=9 atm, V=30 L, and T=301.15K. From our problem, only temperature and pressure changes, while the number of moles, volume and the gas constant, R, stay the same, so they are irrelevant.
Thus, the filled out Combined Gas Law would be:
=
, where the volume, moles of gas, and R are cancelled out.
We can manipulate this equation to derive the new pressure. We find that
9atm≈0.74885.
This means that
≈9/0.74885≈12 atm