D, how many molecules (not moles) of NH3 are produced from 2.05•10^-4 g of H2? u

Question

D, how many molecules (not moles) of NH3 are produced from 2.05•10^-4 g of H2? understand how use stochonety to comertbetweenquantes eeartants and products nchmKaleaatomi balanced reaction by diectly comparing mole as shown in namber el particles between products andreacants O Ask me anything Reaction hydrogen and nitrogen to om ammona NOTE: Throughout thsttoraluse masses euressed toive molar Express your awwer numerically males mol NTla 1.02 10 Incorrect Try Again: 3 attempts remaining Part B cente produced on 230 wolelNa avtencess many grams NHs Express your anwer numerically in grms

Explanation / Answer

3H2 + N2 = 2NH3

Part A:

21.0 Mole H2 *2 Mole NH3 / 3 moles H2= 14 moles NH3

Part B:

2.30 moles N2 * 2 Mole s    NH3 / 1 Moles N2

= 4.6 Mole NH3

Amount of NH3 = 4.6 Mole NH3*17.031 g/ mole

= 78.34 g NH3

3H2 + N2 = 2NH3

There is Part C and D realted to the same question, C how many grams of H2 are needed to produce 10.37g of NH3?

First calculate the number of mole sin 10.37 g NH3:

10.37 g NH3 /17.031 g/ mole

0.61 moles NH3

Now, calculate the moles H2:

0.61 moles NH3 * 3 Moles H2/ 2 moles NH3

= 0.915 Moles H2

Amount of H2: 0.915 Moles H2 *2.01 g/ mole

= 1.84 g

D, how many molecules (not moles) of NH3 are produced from 2.05•10^-4 g of H2?

Number of moles H2= 2.05•10^-4 g /2.01 g/ mole

= 1.02 *10^-4 moles H2

Now moles NH3;

1.02 moles *10^-4 H2 *2 moles NH3/3 Moles H2

6.8*10^-5 Mole NH3

1 moles = 6.023*10^23 molecules

6.8*10^-5 Mole NH3 *6.023*10^23 molecules /1 moles

= 4.1*10^18 Molecules NH3

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