Question

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

3H2(g)+N2(g)→2NH3(g)

a. How many grams of H2 are needed to produce 11.70 g of NH3?

b. How many molecules (not moles) of NH3 are produced from 4.42×10⁻⁴ g of H2?

Answer 1

As the following equation showes

3H₂(g)   +    N₂(g)   →      2NH₃(g)

3(1+1)      (2 x 14)            2(14 +3)

6grams        28 grams     34grams

According to above balanced equation:

34 grams of ammonia (NH₃) gas is produced from 6 grams of hydrogen (H₂) gas.

Therefore 11.70 grams of ammonia (NH₃) gas is produced from = (6 x 11.70) /34

                                                                                              = 2.064 grams

  

Hence, 2.064 grams of hydrogen (H₂) gas is required to produce 11.70 grams of ammonia (NH₃) gas.

1. According to Avogadro’s number 1 mole of gas has 6.023 x 10²³ molecules

So, 2 moles of Ammonia gas will have = 2x 6.023 x 10²³ = 12.046 x 10 ²³ molecules

6 grams of hydrogen gas is used to produce                   = 12.046 x 10 ²³/ 6

                                                                                         = 2.007 x 10 ²³ molecules of (NH₃) gas

4.42 x 10 -4 grams of hydrogen gas is used to produce = (2.007 x 10 ²³) x (4.42 x 10 ^-4)

                                                                                        =8.873 x 10¹⁹ molecules

Hence, 8.873 x 10¹⁹ molecules of NH₃ gas are produced from 4.42 x 10 -4 grams H₂ gas.