Hydrogen gas, H2, reacts with nitrogen gas, N2, to form ammonia gas, NH3, accord

ID: 837422 • Letter: H

Question

Hydrogen gas, H2, reacts with nitrogen gas, N2, to form ammonia gas, NH3, according to the equation

3H2(g)+N2(g)?2NH3(g)

NOTE: Throughout this tutorial use molar masses expressed to five significant figures.

(A)How many moles of NH3 can be produced from 12.0mol of H2 and excess N2?

Express your answer numerically in moles.

8.00

mol NH3

art B

How many grams of NH3 can be produced from 4.10mol of N2 and excess H2.

Express your answer numerically in grams.

Part C

How many grams of H2 are needed to produce 11.35g of NH3?

Express your answer numerically in grams.

Part D

How many molecules (not moles) of NH3 are produced from 3.1810?4g of H2?

Express your answer numerically as the number of molecules.

8.00

mol NH3

3 moles of H2 react with 1 mole of N2 to give 2 moles of NH3

1mole of H2 = 2gm, N2 = 28gm, NH3 = 17gm

in terms of molar masses 6gm of H2 react with 28gm of N2 to give 34gm of NH3

PART - A

3 moles of H2 give 2 moles of NH3

so 12mol of H2 gives 2*4 = 8 moles

PART - B

1 mol of N2 gives 2 mol of NH3

so 4.10mol of N2 gives 2*4.10 = 8.20mol of NH3 = 8.20 *17 = 139.4gm of NH3

PART - C

11.35g of NH3 = 0.667 mol of NH3

3 moles of H2 give 2 moles of NH3

(3*0.667)/2 = 1.0014mol of H2 gives 0.667 mol of NH3

1.0014mol of H2 = 2.0028g H2

PART - D

6gm of H2 gives 2*6.023*10^23 molecules of NH3

3.18*10^-4 grams produces = 3.18*6.023*10^23 /3 = 6.3843*10^19 molecules

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