Answer to Question #141606 in General Chemistry for Javier

Question

When 25.00 mL of 3.42 M Pb(NO3)2 reacts with excess NaCl, how many grams of PbCl2 form?

Pb(NO3)2(aq) + 2 NaCl(aq) → PbCl2(s) + 2 NaNO3(aq)

Answer 1

Mole = concentration × volume

Mole = 3.42M ×( 25 × 10^-3 L)

Mole = 0.0855 mole

1 mole of Pb(NO3)2 = 278.1 g PbCl2

0.0855 mole = 278.1 × 0.0855 g PbCl2

= 23.78g PbCl2