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!

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