Part B How many grams of NH3 can be produced from 3.75 mol of N2 and excess H2 E

Question

Part B How many grams of NH3 can be produced from 3.75 mol of N2 and excess H2 Express your answer numerically in grams View Available Hint(s) g NH3 Submit Pre vious Ans Incorrect; Try Again; 8 attempts remaining Part C How many grams of H2 are needed to produce 14.74 g of NH3? Express your answer numerically in grams View Available Hint(s) Submit Part D How many molecules (not moles) of NH are produced from 5.21x10-4 g of H2? Express your answer numerically as the number of molecules View Available Hint(s)

Explanation / Answer

Part B

Balanced equation:
N2 + 3 H2 ====> 2 NH3
Reaction type: synthesis
Moles of N2 = 3.75 Moles

Moles of NH3 produced =  7.5 Moles

Mass of NH3 produced = 7.5 x 17.03 =  127.72 gm

Part C

14.74 gm of NH3 = 14.74 / 17.03 =  0.8655 Moles

Moles of H2 is needed = 1.29825 x 2.016 =  2.617 gm

Part D

Mass of H2 = 5.21 x 10-4 gm

Moles of H2 =  0.000521 / 2.016 = 2.58 x 10-4 Moles

Moles of ammonia produced = 1.72 x 10-4 Moles

Molecules of Ammonia produced =  1.72 x 10-4 x 6.023 x1023 = 1.037 x 1020 Molecules

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