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?
A.
38.3 g
B.
405 g
C.
52.9 g
D.
476 g
answer is D 476
massHFl=(6molesHFl/2molesNH3) *( 1moleNH3/17gNH3)*(135gNH3)*(19gFl/moleFl)
= (6*1*135*19/2*17)=452 g hydrogen floride
check my work
To determine the mass of hydrogen fluoride gas produced from 135 g of ammonia gas, we will use the balanced equation to calculate the stoichiometry of the reaction.
From the balanced equation:
5 F2 (g) + 2 NH3 (g) -> N2F4 (g) + 6 HF (g)
We can see that for every 2 moles of ammonia gas (NH3) consumed, 6 moles of hydrogen fluoride gas (HF) are produced. This gives us a molar ratio of 2:6 or 1:3.
First, we need to determine the moles of ammonia gas using its molar mass.
Molar mass of NH3 = 14.01 g/mol + (3 x 1.01 g/mol) = 17.03 g/mol
Moles of NH3 = mass / molar mass = 135 g / 17.03 g/mol ≈ 7.92 mol
Since the molar ratio between NH3 and HF is 1:3, the moles of HF produced will be:
Moles of HF = Moles of NH3 x (3/1) = 7.92 mol x 3 = 23.76 mol
Finally, we need to calculate the mass of HF using its molar mass.
Molar mass of HF = 1.01 g/mol + 19.00 g/mol = 20.01 g/mol
Mass of HF = moles of HF x molar mass = 23.76 mol x 20.01 g/mol ≈ 475.7 g
Rounded to the nearest whole number, the mass of hydrogen fluoride gas produced is approximately 476 g.
Therefore, the correct answer is option D. 476 g.
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 ratios between the reactants and products.
First, we need to calculate the number of moles of ammonia gas (NH3) using its molar mass. The molar mass of ammonia (NH3) is calculated as follows:
NH3: 1 nitrogen atom (14.01 g/mol) + 3 hydrogen atoms (3 × 1.01 g/mol) = 17.03 g/mol
Number of moles of ammonia gas = given mass of ammonia gas / molar mass of ammonia gas
Number of moles of ammonia = 135 g / 17.03 g/mol = 7.93 mol
Next, we use the stoichiometry from the balanced equation to determine the number of moles of hydrogen fluoride gas (HF) produced. According to the balanced equation, the molar ratio between NH3 and HF is 2:6.
Number of moles of HF = Number of moles of NH3 × (6 moles HF / 2 moles NH3)
Number of moles of HF = 7.93 mol × (6 mol HF / 2 mol NH3) = 23.79 mol
Finally, to find the mass of hydrogen fluoride gas, we use the molar mass of HF, which is calculated as follows:
HF: 1 hydrogen atom (1.01 g/mol) + 1 fluorine atom (18.99 g/mol) = 20.00 g/mol
Mass of hydrogen fluoride gas = Number of moles of HF × molar mass of HF
Mass of hydrogen fluoride gas = 23.79 mol × 20.00 g/mol = 475.80 g
Rounding to the nearest tenth, the mass of hydrogen fluoride gas produced from 135 g of ammonia gas is 475.8 g.
The correct answer is D. 476 g.