Two atoms of nitrogen form the gaseous, natural state, of nitrogen.
10.62 grams N2 (1 mole N2/28.02 grams)(6.022 X 10^23/1 mole N2)(1 mole N2 atoms/6.022 X 10^23)
0.3790 mole of gaseous nitrogen atoms
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* as you may see, Avogadro's number is over itself as a form of one and is a superfluous step put there for formality's sake
3-2
3.2
The volume occupied by 3 moles of nitrogen gas will be different, depending on the temperature and pressure of the gas.
molar weight of N = 14 grams/mole 35.7 grams/14 grams/mole = 2.55 moles
N2 + 3H2 --> 2NH3 You have been told, indirectly, that nitrogen limits and will drive the reaction. 3 moles N2 (2 moles NH3/1 mole N2) = 6 moles ammonia gas produced ========================
Assuming ideal behaviour, 1 mole of any gas occupies 22.4L at STP. So, moles of 10L = 10/22.4 moles = 0.4464 moles
.271 mol
The volume occupied by 3 moles of nitrogen gas will be different, depending on the temperature and pressure of the gas.
molar weight of N = 14 grams/mole 35.7 grams/14 grams/mole = 2.55 moles
3.6
N2 + 3H2 --> 2NH3 You have been told, indirectly, that nitrogen limits and will drive the reaction. 3 moles N2 (2 moles NH3/1 mole N2) = 6 moles ammonia gas produced ========================
9H2 + 3N2 ------> 6NH3 3 moles of N2 would be required.
Each of the nitrogen atoms in N2 needs three electrons, for a total of six.
Assuming ideal behaviour, 1 mole of any gas occupies 22.4L at STP. So, moles of 10L = 10/22.4 moles = 0.4464 moles
To form ammonia, balanced reaction is N(2) + 3H(2) ---> 2NH(3) + H(2)O. As you can see for 1 mole of nitrogen three moles of hydrogen is required. Hence for your question, 3 moles nitrogen is required to satisfy the ratio.
.271 mol
The chemical equation is N2 + 3H2 -> 2NH3 So reacting 2 moles of N2 will produce 4 moles ammonia.
PV = nRT ⟹ n = PV/RT = 1 * 18.65 / (0.082 * 273.15) = 0.8321 moles.
That amount of ammonia contains two moles of hydrogen gas. One mole of hydrogen gas weighs 2.016 grams. Therfore 3.75 grams of ammonia contains two moles of hydrogen.