Question

A student added 10g of 10%lead nitrate to 10 gram of 10% sodium chloride. after the reaction was complete, he found 18 gram of solution .what was the amount of lead chloride formed?(could someone please explain it's urgent I got my exam tomorrow!)

Answer 1

Pb (NO3) 2 + 2NaCl--------> PbCl2 + 2 NaNO3.

Molecular Weight of Pb (NO3) 2 = 331 g / mol

Molecular weight of NaCl = 58.5 g / mol

So 10 g of Pb (NO3) 2 will react with

=10 x 58.5 / 331 g of NaCl = 1.75 g of NaCl.

The amount of PbCl2 formed = 10 + 1.75 = 11.75 g.

The rest of ( 10 – 1.75 ) = 8.25 g of NaCl, will remain unreacted.

Answer 2

Hi! I hope this could help you in your exam.