How do you balance Pb(NO3)4+K2CrO4-> Pb(CrO4)2+KNO3?

Answer 1

To balance this equation you need to look at how many of a element is on one side
On the right side you need a four in front of the KNO3 on the left sode you need to put a 2 in front of the K2CrO4

Answer 2

Pb(NO3)4+K2CrO4-> Pb(CrO4)2+KNO3

you have 4 NO3 on the left, so you need to have 4 NO3 on the right. How? Add a 4 in front of KNO3.

Pb(NO3)4+K2CrO4-> Pb(CrO4)2+4 KNO3

Now you have 4 K on the right, but only 2 K on the left. So add a 2 in front of K2CrO4.

Pb(NO3)4 + 2 K2CrO4-> Pb(CrO4)2 + 4 KNO3

Now, you have one Pb on the left, one Pb on the right
4 NO3 on the left, 4 NO3 on the right
4 K on the left, 4 K on the right
2 CrO4 on the left, 2 CrO4 on the right

it's balanced!

Answer 3

First look at the charges: Pb is +4, NO3 is -1 (and there are 4 NO3) so it is -4, K2 is +2 (2 *+1 =+2) and CrO4 is -2. So find a common factor (4) and multiply everyone by the number they need to become 4 (ignore the + and-): PB and (NO3)4 are multiplied by 1 because 1 *4=4 and K2 and (CrO4) are multiplied by 2 because 2*2=4 and voila the rxn eqn becomes:
Pb(NO3)4 + 2K2(CrO4)>>PB(CrO4)2 + 4KNO3

Answer 4

Weeelll.... you have to ask yourself if anything "funny" is going on first.. is this a redox reaction (reduction/oxidation taking place)?..
Tips to consider:
The lead salt on the reactants side: what is the oxidation state of the lead...+4..maybe...???
The lead salt on the products side: what is the oxidation state...still +4.....??
Your chromium in the chromium salt isn't changing oxidation state either...maybe this is an example of a double displacement reaction..perhaps.. so it's just a matter of balancing the mole amounts of 4 things really: 1) Pb, 2) NO3, 3) K , and 4) CrO4, just make the mol amounts of 1), 2), 3), and 4) equal on both sides of the chemical equation... not that difficult really

Answer 5

Pb(NO3)4 + 2K2CrO4 ---> Pb(CrO4)2 + 4KNO3

I'm pretty sure that's right. Hope I helped