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.
I believe that it is petroleum ether.
Ammonium chloride is the correct name for NH4CL
Some minerals tend to look alike.
207.217 amu
Work:
203.973 amu *(0.014) = 2.855 amu
205.974 amu *(0.241) = 49.639 amu
206.976 amu *(0.221) = 45.741 amu
207.977 amu *(0.524) = 108.979 amu
2.855 + 49.639 + 45.741 + 108.979 = 207.217 amu