Answer to Question #105006 in General Chemistry for Kaci

Question

How many grams of lead nitrate are needed to produce 105.25 g of sodium nitrate?

Answer 1

$Pb(NO_3)_2 + 2NaOH = 2NaNO_3 + Pb(OH) _2$

$\nu(NaNO_3) = m/M = 105.25/85 = 1.24 mol.$

1.24 : 2 = $\nu(Pb(NO_3)_2)$ : 1

$\nu(Pb(NO_3)_2)=1.24/2=0.62 mol.$ $m(Pb(NO_3)_2)=\nu*M=0.62*331=205.22 g.$

Answer: 205.22 g.