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# Answer to Question #127447 in Inorganic Chemistry for aleeyah

Question #127447
Part B
How many grams of NH3 can be produced from 2.49 mol of N2 and excess H2.

___________g NH3

Part C
How many grams of H2 are needed to produce 11.17 g of NH3?

_________g H2

Part D
How many molecules (not moles) of NH3 are produced from 5.20Ã—10âˆ’4 g of H2?

_____molecules
1
2020-07-27T05:36:21-0400

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 * 1020. ) . Answer .

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