molar weight of N = 14 grams/mole
35.7 grams/14 grams/mole = 2.55 moles
The volume occupied by 3 moles of nitrogen gas will be different, depending on the temperature and pressure of the gas.
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 chemical equation is N2 + 3H2 -> 2NH3 So reacting 2 moles of N2 will produce 4 moles ammonia.
The volume occupied by 3 moles of nitrogen gas will be different, depending on the temperature and pressure of the gas.
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.
3.50 W 21.80 x 1/14th