Question

Ammonia gas reacts with chlorine trifluoride gas to produce nitrogen gas, chlorine gas, and hydrogen fluoride gas. if 30.0 g of ammonia reacts with 175 g of chlorine trifluoride, what mass (in g) of hydrogen fluoride gas is produced?

Answer 1

The mass of hydrogen fluoride gas produced is 113.6 g.

Step-by-step explanation:

The balanced chemical equation for the reaction between ammonia gas and chlorine trifluoride gas is:

8 NH3 (g) + 3 ClF3 (g) → 6 NF3 (g) + 9 HCl (g)

To find the mass of hydrogen fluoride gas produced, we need to calculate the amount of chlorine trifluoride that reacts completely with 30.0 g of ammonia. We can start by determining the moles of chlorine trifluoride:

175 g ClF3 * (1 mol ClF3 / 92.45 g ClF3) = 1.892 mol ClF3

Since the stoichiometric ratio between ClF3 and HF is 3:9, we can use this ratio to calculate the moles of HF:

1.892 mol ClF3 * (9 mol HF / 3 mol ClF3) = 5.676 mol HF

Finally, we can convert the moles of HF to grams:

5.676 mol HF * (20.01 g HF / 1 mol HF) = 113.6 g HF