How many molecules (not moles) of NH3NH3 are produced from 6.09*10-4 gg of H2H2? Express your answer numerically as the number of molecules.

Answer: 1.22x10^20 molecules

Explanation:

N2 + 3H2 - > 2NH3

Molar Mass of H2 = 2g/mol

Mass conc of H2 = 6.09*10^-4

n = Mass conc / molar Mass

n = 6.09*10^-4 / 2 = 3.045x10^-4mol

From the equation,

3 moles of H2 produced 2moles of NH3.

Therefore, 3.045x10^-4mol of H2 will be produce = (3.045x10^-4x2) / 3

= 2.03x10^-4mol of NH3

1mole of NH3 contains 6.02x10^23 molecules.

Therefore, 2.03x10^-4mol of NH3 will contain = (6.02x10^23x2.03x10^-4) = 1.22x10^20 molecules

Molecules NH3 = 1.21*10^20 molecules

Explanation:

Step 1: Data given

Mass of H2 = 6.09 * 10^-4 grams

Molar mass of H2 = 2.02 g/mol

Step 2: The balanced equation

N2 + 3H2 → 2NH3

Step 3: Calculate moles H2

Moles H2 = mass H2 / molar mass H2

Moles H2 = 6.09 * 10^-4 grams / 2.02 g/mol

Moles H2 = 3.01*10^-4 moles

Step 4: Calculate moles NH3

For 1 mol N2 we need 3 moles H2 to produce 2 moles NH3

For 3.01 * 10^-4 moles H2 we'll have 2/3 * 3.01*10^-4 = 2.01*10^-4 moles

Step 5: Calculate molecules NH3

Molecules NH3 = 2.01*10^-4 moles * 6.022 * 10^23

Molecules NH3 = 1.21*10^20molecules