Question

4NH3+5O2---->4NO+6H2O
a.) if you start with 80 grams of o2 and 70 grams of NH3, how many grams of NO will you produce?
b.) How many grams of h2o will you make?
Please show your work or explain this is from a test

Answer 1

Answer:

Explanation:

The reaction expression is given as:

          4NH₃    +     5O₂    →    4NO   +   6H₂O

Given parameters:

   Mass of O₂  = 80g

   Mass of NH₃  = 70g

a. how many grams of H₂O is produced.

To solve this problem, we need to find the limiting reactant. This is the reactant in short supply and it will determine the extent of the reaction.

 Convert the masses to number of moles;

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

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

 Molar mass of O₂   = 2(16)  = 32g/mol

     Number of moles of NH₃   = $\frac{17}{70}$   = 0.243mole

     Number of moles of O₂   = $\frac{32}{80}$   = 0.4mole

From the balanced reaction equation:

            4 moles of NH₃  will react with 5 moles of O₂  

            0.243 mole of NH₃ will react with $\frac{0.243 x 5}{4}$   = 0.3 mole of O₂

Here we clearly see that ammonia is in short supply.

So;

     if 4 moles of NH₃ produced 4moles of NO;

        0.243 moles of NH₃ will produce 0.243 mole of NO

  Mass of  = number of moles x molar mass

    Molar mass of NO = 14 + 16 = 30g/mol

  Mass of NO = 0.243 x 30  = 7.29g

b. Mass of H₂O;

   From the balanced reaction equation:

        4moles of NH₃  will produce 6 mole of H₂O

   0.243 mole of NH₃ will therefore produce $\frac{0.243 x 6}{4}$   = 0.365mole of H₂O

 Molar mass of H₂O  = 2(1)  + 16  = 18g/mol

Mass of H₂O  = 0.365 x  18  = 6.56g