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When the reactants are
Lead nitrate formula pb(NO3)2
Potassium Chromate formula K2CrO4

A explanation will be useful for this reaction.

Thanks, this is part of GCSE Science Coursework.

2007-03-04 01:15:43 · 4 answers · asked by Anonymous in Science & Mathematics Chemistry

4 answers

Pb(NO3)2 + K2CrO4 ---> Pb++ + 2 NO3- + 2 K+ + CrO4-- in aqueous solution (ie. just forming the ions)

I guess they would recombine to form some Pb(NO3)2 and some PbCrO4 and some KNO3 and K2CrO4 again

but ooooh look PbCrO4 is a yellow solid insoluable in water (see the wiki page) so you get


--------> PbCrO4(s) + 2 KNO3(aq)

where the solid precipitates out and leaves the K+ and NO3 - in solution.

Turns out Lead Nitrate is one of the very few soluable lead salts and a favourite for demonstrating precipitation (forms many insoluable lead salts which are often strikingly coloured) and was used in the formation of many pigments of the yellow/orange variety.

2007-03-04 01:36:16 · answer #1 · answered by Orinoco 7 · 0 0

Pb(NO3)2 + K2CrO4 = 2KNO3 + PbCrO4
All nitrates and all alkali metal salts are soluble.
So the two reactants are soluble.
The product 2KNO3, which contains both an alkali metal and a nitrate remain in solution.
However, the Lead Chromate(PbCrO4) is insoluble and will precipitate out - to the bottom of the reaction vessel.

2007-03-04 07:29:26 · answer #2 · answered by lenpol7 7 · 0 0

This looks like a double replacement reaction
so first...
Pb(NO3)2 + K2CrO4

then switch the elements and the polyatomic ions to get new substances with new formulas...
PbCrO4 + KNO3

so the whole balanced equation is lead(II) nitrate plus potassium chromate yeilds lead(II) Chromate and potassium nitrate.
The equation...
Pb(NO3)2 + K2CrO4 --> PbCrO4 + KNO3

then balanced...
Pb(NO3)2 + K2CrO4 --> PbCrO4 + 2 KNO3

2007-03-04 01:34:08 · answer #3 · answered by Kurt 2 · 0 0

I am only 14 years old, but i am doing int2 chemistry at school and i think it might be:
Pb(NO3)2 + K2CrO4 ---> PbCrO4 + 2 KNO3

Hope that helps you!

Cj

2007-03-04 02:54:56 · answer #4 · answered by Cj 1 · 0 0

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