How do you balance K2Cr2O7 + AgNO3?

Question

Caleb Adams · Accepted Answer

#K_2Cr_2O_7 + 2AgNO_3 = Ag_2Cr_2O_7 + 2KNO_3#

I'm going to assume that the above equation is true because you did not include the final product in your question.

There are two ways to balance this: either count each atom separately, or apply your knowledge of ions.

I'm going to use step 2 because step (1) in this type of equation would be much more difficult.

First, rewrite the equation as follows to prevent yourself from becoming confused:

#(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.

Consequently,

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^"-")#

Going back to the first format, you can now write the equation for balance as

#K_2Cr_2O_7 + 2AgNO_3 = Ag_2Cr_2O_7 + 2KNO_3#

Peyton Adams · Answer

undefined 2 K2Cr2O7 + 14 AgNO3 → 2 K2Cr(NO3)6 + 3 Ag2Cr2O7 + 6 KNO3