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.

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

it have destructive character

if the concentration of products is decreased, the equilibrium will shift toward the products). add an inert gas (one that is not involved in the reaction) to the constant-volume reaction mixture: this will increase the total pressure of the system, but will have no effect on the equilibrium condition.

explanation: