Question

N₂ + 3H₂>>>2NH₃

Using the equation above, calculate how many grams of NH₃ you would have if you started with 2.37 g of N₂ and 2.37 g of H₂?

Answer 1

First , we have to calculate the limiting reagent.

            N₂   +  3H_{2       --------------->}   2NH₃

    moles of N2 = 2.37 g / 28 g/mol = 0.085 mol

   moles of H2 = 2.37 g / 2 g/mol = 1.185 mol

     From above equation,

      1 mol of N2 requires 3 moles of H2.

So, 0.085 mol of N2 requires 3 x 0.085 mol = 0.255 mol of H2.

But, we have 1.185 mol of H2.

So, we have excess H2.

Hence, N2 is the limiting reagent.

    Yield is calculated based on N2.

                N₂                        +      3H_{2       --------------->}   2NH₃

          1 mol = 28 g                                        2 mol = 2x 17 g = 34 g        

             2.37 g                                                           ?

                    ? = ( 2.37 g / 28 g) x 34 g NH3

                      = 2.87 g of NH3

       Therefore,

         yield of NH3 = 2.87 grams