1) Given the following chemical equation, how many grams of N2 are produced by 8

Question

1) Given the following chemical equation, how many grams of N2 are produced by 8.39 g of H2O2 and 6.21 g of N2H4.

2H202(l) + N2H4 (l) --> 4H2O(g) + N2 (g)

2) Nitrogen and hydrogen combine at high temperature, in the presence of a catalyst, to produce ammonia. This is the formula: N2(g) + 3H2(g) ---> 2NH3(g)

Assume 0.260 mol of N2 and 0.865 mol of H2 are present intially. After complete reaction, how many moles of amonia are produced? How many moles of H2 and N2 remain? What is the limiting reactant?

Explanation / Answer

1)    2H202(l) + N2H4 (l) --> 4H2O(g) + N2 (g)

      2*34           32                                 28

68 g of H202 wil react with 32 g of N2H4

8.39 g of H2O2 will react with =(32/68)*8.39=3.95 g of N2H4

but 6.21 g of N2H4 is present .so N2H4 is excess

68 g of H202 wil produce with 28 g of N2

8.39 g of H2O2 will produce=(28/68)*8.39

                                           =3.455 g of N2

2) N2(g) + 3H2(g) ---> 2NH3(g)

    1 mole   3 moles

1 mol of N2 reacts with 3 moles of H2

0.260 mol of N2 reacts =(3/1)*0.26=0.78

but 0.865 mol of H2 are present so H2 is excess reactant

N2 is limiting reactant.

1 mol of N2 forms 2 moles of NH3

0.260 mol of N2 forms =(2/1)*0.26=0.52 moles of NH3

moles od NH3=0.52 moles
moles od N2 remiang=0.26-0.26=0 moles

moles od H2 remiang=intial moles of H2 - reacted moles of H2

                                 =0.865-0.78=0.085 moles

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