Answer to Question #143985 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

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

Mole ratio is 1:1

Moles of Pb(NO3)2 produces = (M X V)/1000

therefore, moles = (3.42 x 0.025 )/1000 = 0.0855 moles

On that note, 0.0855 moles of PbCl2 are produced.

But moles = mass/RFM

thus, mass produces = moles x RFM

= 0.0855 X 278.106

= 23.78g