Fluorine gas can react with ammonia gas to produce dinitrogen tetrafluoride gas and hydrogen fluoride gas. This reaction is described by the following balanced equation:

5 F2 (g) + 2 NH3 (g) N2F4 (g) + 6 HF (g)

What mass of hydrogen fluoride gas is produced from 135 g of ammonia gas assuming that there is an ample amount of fluorine gas present for the reaction to be completed?

mols NH3 = grams/molar mass

Convert mols NH3 to mols HF using the coefficients in the balanced equation.
Convert mols HF to grams HF. g = mols x molar mass.

To determine the mass of hydrogen fluoride gas produced from 135 g of ammonia gas, we need to use the balanced equation and the molar masses of ammonia (NH₃) and hydrogen fluoride (HF) gas.

The balanced equation tells us that the ratio of ammonia to hydrogen fluoride is 2:6. This means that for every 2 moles of ammonia gas, we will produce 6 moles of hydrogen fluoride gas.

1. Calculate the number of moles of ammonia gas:
Moles of NH₃ = Mass of NH₃ / Molar mass of NH₃

The molar mass of NH₃ is calculated by adding up the atomic masses of nitrogen and hydrogen:
Molar mass of NH₃ = 1(14.01 g/mol) + 3(1.01 g/mol) = 17.03 g/mol

Moles of NH₃ = 135 g / 17.03 g/mol

2. Use the mole ratio from the balanced equation to calculate moles of HF:
Moles of HF = (Moles of NH₃) × (Mole ratio of HF/NH₃)

The mole ratio of HF to NH₃ is 6/2 = 3.

Moles of HF = (Moles of NH₃) × 3

3. Calculate the mass of HF using the moles of HF:
Mass of HF = Moles of HF × Molar mass of HF

The molar mass of HF is calculated by adding up the atomic masses of hydrogen and fluorine:
Molar mass of HF = 1(1.01 g/mol) + 1(19.00 g/mol) = 20.01 g/mol

Mass of HF = (Moles of HF) × (Molar mass of HF)

Now, let's calculate the values:

Moles of NH₃ = 135 g / 17.03 g/mol = 7.933 mol
Moles of HF = (7.933 mol) × 3 = 23.799 mol
Mass of HF = (23.799 mol) × (20.01 g/mol) = 476.45 g

Therefore, the mass of hydrogen fluoride gas produced from 135 g of ammonia gas is approximately 476.45 g.

To determine the mass of hydrogen fluoride gas produced, we need to use the balanced equation and the molar mass of ammonia gas (NH3).

Let's begin by finding the molar mass of NH3:
N: 1 atom x 14.01 g/mol = 14.01 g/mol
H: 3 atoms x 1.01 g/mol = 3.03 g/mol
Total molar mass of NH3 = 14.01 g/mol + 3.03 g/mol = 17.04 g/mol

Now, let's calculate the number of moles of NH3 in 135 g:
Number of moles = mass / molar mass
Number of moles of NH3 = 135 g / 17.04 g/mol ≈ 7.92 mol

From the balanced equation, we see that the molar ratio between NH3 and HF is 2:6. It means that for every 2 moles of NH3, 6 moles of HF are produced.

So, using the molar ratio, we can calculate the number of moles of HF produced:
Number of moles of HF = (7.92 mol NH3) x (6 mol HF / 2 mol NH3) = 7.92 mol x 3 = 23.76 mol

Finally, let's calculate the mass of HF produced using its molar mass:
Molar mass of HF: 1 atom of H = 1.01 g/mol ; 1 atom of F = 19.00 g/mol
Total molar mass of HF = 1.01 g/mol + 19.00 g/mol = 20.01 g/mol

Mass of HF = number of moles x molar mass
Mass of HF = 23.76 mol x 20.01 g/mol ≈ 475.26 g

Therefore, approximately 475.26 grams of hydrogen fluoride gas are produced from 135 grams of ammonia gas.