Question

N2H4 can be oxidized to N2 by K2CrO4 which is reduced to Cr(OH)4^-. Write the ionic equation eliminate Spector ions and then write the chemical equation in basic conditions.How many grams of N2H4 can be oxidized when 24.0g K2CrO4 is used?

Answer 1

N2H4 changes to N2 by losing 4 electrons and CrO4-2 changes to Cr(OH)4- by gaining 3 electrons

4K2CrO4+ 3 N2H4 + 4H2O -----------> 4KCr(OH)4+ 3N2 + 4KOH

thus the ionic equation is

8K+ + 4CrO4^-2 + 3 N2H4 + 4H2O -----------> 8K+ + 4Cr(OH)4^- + 3N2 + 4OH^-  

Thus the spectator ions are only K+ ions

the net ionic equation is

4CrO4^-2 + 3 N2H4 + 4H2O -----------> 4Cr(OH)4^- + 3N2 + 4OH^-  

molar mass of potassium dichromate = 194.19 g/mol

molar mass of N2H4 = 32g/mol

3 mole of N2H4 can be oxidised by 4 moles of chromate

Thus moles of N2H4 that can be oxidised by 24/194.19 moles of chromate

= 24x3/[194.19x4]

=0.0927 moles

Thus mass of N2H4 that can be oxidised = moles x molar mass

= 0.0927 molx 32g/mol

= 2.966 g