Question

!!!!!!!!!!!!!!!!!! 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

Answer 1

A. 828.03 grams

B. 54.54 grams

C. 218.14 grams

D. 3312.10 grams

E. 110.00 grams

Answer 2

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!