In the following reaction, how many grams of lead(ii) nitrate pb(no3)2 will produce 425 grams of sodium nitrate (nano3)?
pb(no3)2(aq) + 2 nabr(aq) pbbr2(s) + 2 nano3(aq)

the molar mass of pb(no3)2 is 331.21 grams and that of nano3 is 85 grams.
a) 828.03 g b) 54.54 g c) 218.14 d) 3312.10 g e) 110.00 g

The balanced reaction as you have stated is:
Pb(NO3)2 (aq) + 2NaBr (aq) --> PbBr2 (s) + 2NaNO3 (aq)
Therefore for every mole of Pb(NO3)2 reacted, 2 moles of NaNO3 is produced.
The number of moles of NaNO3 produced can be calculated as:
moles = mass / relative molecular mass = 425 g / 85.00 g/mol = 5.00 mol
Therefore the number of moles of Pb(NO3)2 is 3.90 / 2 = 2.50 mol
This can be used to calculate the mass of Pb(NO3)2 used:
mass = moles * RMM = 2.50 mol * 331.21 g/mol = 828.03 g.
Therefore the answer is A.

Hope this helps!

The density of water is 1 g/cm cubed and in order for the pebble to sink its density has to be greater than the density of water and in order for the twig to float its density has to be less than the density of water

c. electrical charges of ionic bonds.

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