N2 + 3h2 = 2nh3

for the equation in (2) above, you have 1000. g of n2 and 150.0 g of h2:

a. identify the limiting reactant.

b. calculate the theoretical yield in grams of nh3.

c. calculate the grams of excess reactant.
show work

The answer to your question is:

a) Hydrogen

b) 850 g of NH3

c) 300 g of N2

Explanation:

                                       N2    +     3H2    ⇒    2NH3

N2 = 1000 g

H2 = 150 g

a. Identify the limiting reactant.

MW N2 = 28 g

MW H2 = 2 g        

MW NH3 = 17g                            

                                    N2    +     3H2    ⇒    2NH3

                                   28g  3(2)

                               1000 g   x

                              x = (1000 x 6)/ 28

                             x = 214.3 g of H2

Then, the limiting reactant is H2, because is necessary to have 214.3 g of H2 to react with 1000 g of N2, and there are only 150 g.

b. Calculate the theoretical yield in grams of NH3.

                                  N2    +     3H2    ⇒    2NH3                                

                                                    6g 2(17)g

                                               150g   x

                                              x = (150 x 34) / 6

                                              x = 850 g of NH3

c. Calculate the grams of excess reactant.

                                  N2    +     3H2    ⇒    2NH3

                                 28g  6 g

                                  x 150 g

                                 x = (150 x 28)/ 6

                                 x = 700 g of N2

Excess reactant = 1000g  - 700g

                          = 300 g of N2

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the enthalpy of formation of water is given as -285.8kj/mol, which states that the enthalpy of formation is negative.

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the formation of water takes place by giving out heat which is -285.8 kj/mol.