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.
828.03 grams
54.54 grams
218.14 grams
3312.10 grams
110.00 grams

Answer 1

m1= 425 grams of sodium nitrate (NaNO3)
m2= ? grams of lead(II) nitrate Pb(NO3)2
MM1= 85 grams of NaNO3 is
MM2=molar mass of Pb(NO3)2 is 331.21 grams

1) convert the grams of sodium nitrate to moles of sodium nitrate ( n1=m1/MM1) n=moles: m1=mass of sodium nitrate; MM1= molecular mass of sodium nitrate.
2) according to the stoiquiometry of the reaction you can see that 1 mol of Pb(NO3)2 will produce 2 mol of NaNO3(aq). Then divide n1 by 2 to calculate the n2 or moles of Pb(NO3)2 that you will need. n2= n1/2
3) convert the moles of Pb(NO3)2 to grams with the same formula used in 1) but now isolate m. => m2= n2/MM2