Answer to Question #127447 in Inorganic Chemistry for aleeyah

Question

Part B
How many grams of NH3 can be produced from 2.49 mol of N2 and excess H2.
Express your answer numerically in grams.

___________g NH3


Part C
How many grams of H2 are needed to produce 11.17 g of NH3?
Express your answer numerically in grams.

_________g H2

Part D
How many molecules (not moles) of NH3 are produced from 5.20×10−4 g of H2?
Express your answer numerically as the number of molecules.

_____molecules

Answer 1

part B .

N₂ + 3 H₂ ---------> 2 NH₃ . Is The Reaction .

 1 mole of N₂ Gives 2Mole of NH₃ .

So , 2.49 mole of N₂ gives(2* 2.49 ) mol of NH₃ .

So , Mass Of NH₃ Formed = 2* 2.49 * 17 = 84.66 g .

Part C .

11.17 g of NH₃ have number of moles of NH₃ = 11.17 / 17 = 0.657 mol.

N2 + 3 H2 ---------> 2 NH3 . Is The Reaction .

2 Mole of NH₃ formed by 3 mol of H₂ .

moles of H₂ Required = 3* 0.657 / 2 . = 0.9855 moles.

Mass of H₂ = 0.9855 mol * 2 g / mol . = 1.971 g .

part D

5.20×10^-4 g of H₂ have moles of H₂ = 5.20 * 10^-4 / 2 = 2.6 * 10^-4 mole . of H₂ .

N2 + 3 H2 ---------> 2 NH3 . Is The Reaction

3 mol of H₂ GIVES 2 mole of NH₃

So , 2.6 * 10^-4 mol of H₂ Gives 2 * (2.6 * 10^-4) / 3 mol of NH₃ Formed .

So , Moles Of NH₃ Formed = 1.73333 * 10^-4 mol of NH₃ formed .

number of molecules of NH₃ = ( moles of NH₃ ) * ( 6.02 * 10²³ ) =

= ( 1.73333 * 10^-4 * 6.02 * 10²³ )

= ( 10.43465 * 10¹⁹ .).

= ( 1.043465 * 10^20. ) . Answer .