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!

Spontaneity tell us the direction of the reaction, but not how fast it goes. gibbs free energy brings together enthalpy, entropy and temperature in one easy formula. g = h - ts. and, finally the sign of delta g is always negative for a spontaneous reaction.

The chemical bubbles