WILL GIVE BRAINLIEST
2 answers:
Answer:
about 1,232 L
Explanation:
Okay! I wasn't sure if you still needed an answer, but I figured I'd throw my two sense in.
To answers this question, you have to know that there are about 1 mole of 02 per 22.4 liters.
Knowing this, we use stoichiometry to figure out how many moles of o2 react with c3h8.
So we have 55 moles of o2!
So then we just convert 55 moles into liters.
55 moles * 22.4L/1 mole o2= 1,232L
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We have to get the amount of nitrogen to be consumed to get 0.75 moles of ammonia.
The amount of nitrogen (in grams) required to prepare 0.75 moles of ammonia is: 10.5 grams.
Ammonia (NH₃) can be prepared from nitrogen (N₂) as per following balanced chemical reaction-
N₂ (g) + 3H₂ (g) ⇄ 2NH₃ (g)
According to the above reaction, to prepare 2 moles of ammonia, one mole of nitrogen is required. Hence, to prepare 0.75 moles of ammonia, moles = 0.375 moles of nitrogen is required.
Molar mass of nitrogen is 28 grams, i.e, mass of one mole of nitrogen is 28 grams, so mass of 0.375 moles of nitrogen is 0.375 X 28 grams=10.5 grams of nitrogen.
Therefore, the amount of nitrogen (in grams) required to prepare 0.75 moles of ammonia is 10.5 grams.