If .75 moles of ammonia is needed, how many grams of nitrogen will be consumed?
1 answer:
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.
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Answer : The metal used was iron (the specific heat capacity is ).
Explanation :
In this problem we assumed that heat given by the hot body is equal to the heat taken by the cold body.
where,
= specific heat of unknown metal = ?
= specific heat of water =
= mass of unknown metal = 150 g
= mass of water = 200 g
= final temperature of water =
= initial temperature of unknown metal =
= initial temperature of water =
Now put all the given values in the above formula, we get
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Therefore, the metal used was iron (the specific heat capacity is ).