How do you solve #K_2PtCl_4 + NH_3 -> Pt(NH_3)_2Cl_2 + KCl#?

1 Answer
Jun 25, 2018

Answer:

#K_2PtCl_4# #+ 2# #NH_3# #-># #Pt(NH_3)_2Cl_2# #+ 2# #KCl#

Explanation:

Start by counting the atoms on both sides of the equation:

L - Atom - R
2 - K - 1
1 - Pt - 1
4 - Cl - 3
1 - N - 2
3 - H - 6

Upon inspection, looks like only Pt is balanced on both sides ( L eft and R ight).

Start by placing a 2 in front of #KCl# since there are 2 K on the left, but only 1 K on the right. Doing so will balance out K and Cl.

Next, place a 2 in front of #NH_3# since there is 1 N on the left and 2 N on the right & 3 H on left and 6 H on right. Doing so will balance out N and H.

These 2 steps will balance the entire equation:
#K_2PtCl_4# #+ 2# #NH_3# #-># #Pt(NH_3)_2Cl_2# #+ 2# #KCl#

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