How do you balance K2Cr2O7 + AgNO3?

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Answer 1 · 2015-10-24T09:07:23

#K_2Cr_2O_7 + 2AgNO_3 = Ag_2Cr_2O_7 + 2KNO_3#

Since you did not put the end product in your question, I'm going to assume that the above equation is true.

How do you balance this? There are two ways: (1) you can separately tally each atom or (2) you can use your knowledge of ions.

Since it would be much more complicated to do step (1) in this kind of equation, I'm going to go with step 2.

First, to avoid confusing yourself, rewrite the equation into this:

#(2K^"+" + Cr_2O_7^"2-") + (Ag^"+" + NO_3^"-") = Ag_2Cr_2O_7 (s) + (K^"+" + NO_3^"-")#

Left side:
#K^+# = 2 (based on initial subscript)
#Cr_2O_7^"2-"# = 1
#Ag^+# = 1
#NO_3^-# = 1

Right side:
#K^+# = 1
#Cr_2O_7^"2-"# = 1
#Ag^+# = 2
#NO_3^-# = 1

Notice that the #K^+# and #Ag^+# ions are not balanced. To balance, you need to multiply both ions by 2.

Thus,

Left side:
#K^+# = 2
#Cr_2O_7^"2-"# = 1
#Ag^+# = 1 x 2 = 2
#NO_3^-# = 1 x 2 = 2

Right side:
#K^+# = 1 x 2 = 2
#Cr_2O_7^"2-"# = 1
#Ag^+# = 2
#NO_3^-# = 1 x 2 = 2

#(2K^"+" + Cr_2O_7^"2-") + 2(Ag^"+" + 2NO_3^"-") = Ag_2Cr_2O_7 (s) + 2(K^"+" + NO_3^"-")#

Reverting back to the original form, now you can write the balanced equation as

#K_2Cr_2O_7 + 2AgNO_3 = Ag_2Cr_2O_7 + 2KNO_3#