Answer:
476g of hydrogen fluoride was produced
Explanation:
In this question, we have 2 moles of ammonia gas producing 6 moles of hydrogen Fluoride gas. This is the theoretical output we have. Let’s try and get what actually is happening!
Firstly, we calculate the number of moles of ammonia in 135g of ammonia.
Mathematically, this can be calculated by dividing the mass given by the molar mass of ammonia.
For NH3, the molar mass is 14 + 3(1) = 17g/mol where 14 and 1 are the atomic masses of nitrogen and hydrogen respectively.
The number of moles of ammonia reacted is thus 135g/17g/mol = 7.94 moles of ammonia
Theoretically, 2 moles ammonia yielded 6 moles hydrogen Fluoride, thus 7.94 moles of ammonia will yield = (7.94 * 6)/2 = 23.82 moles of Hydrogen Fluoride
we now proceed to calculate the equivalent mass of HF here.
To calculate this, we simply multiply the number of moles of hydrogen fluoride by the molar mass of hydrogen fluoride
The molar mass of hydrogen fluoride HF is 1 + 19 = 20g/mol
where 1 and 19 are the atomic masses of hydrogen and fluorine respectively.
The mass produced is thus 20 * 23.82 = 476.47 which is approximately 476g to the nearest whole number
