Question

How many grams of NH3 are required to produce 4.65 g of HF

Answer 1

Answer:

1.32g

Explanation:

Given parameters:

Mass of HF   = 4.65g

Unknown:

Mass of NH₃ = ?

Solution:

We need to know the reaction equation first of all. The reaction that will produce HF involves;

                 2NH₃    +    5F₂    →    N₂F₄   + 6HF

Now that we know the reaction equation;

Find the number of moles of the HF;

        Number of moles  = $\frac{mass}{molar mass}$

         Molar mass of HF  = 1 + 19 = 20g/mol

        Number of moles  = $\frac{4.65}{20}$   = 0.23mole

From the balanced reaction equation;

        6 moles of HF is produced from 2 moles of NH₃  

        0.23 moles of HF will be produced from  $\frac{0.23 x 2}{6}$    = 0.08moles

Now we find the mass of NH₃

          Mass of NH₃ = number of moles x molar mass

                   Molar mass of NH₃  = 14 + 3(1) = 17g/mol

         Mass of NH₃   = 0.08 x 17  = 1.32g