Determine the mass of lead chloride precipitated when 5g of sodium chloride solution reacts with lead trioxonitrate v.

Answer 1

The balanced equation for this reaction is as follows:

2NaCl + Pb(NO3)2 ---> 2NaNO3 + Pb(Cl2)

So, 2 moles of sodium chloride react to form 1 mol of lead chloride.

No. of moles of NaCl = m/M.M
= 5/58.4
= 0.085

So, 0.0425 moles of PbCl2 are formed.

Mass of PbCl2 = n * M.M
= 0.0425 * 271.8
= 11.55g

Answer 2

The balanced equation for this reaction is as follows:

2NaCl + Pb(NO3)2 ---> 2NaNO3 + Pb(Cl2)

So, 2 moles of sodium chloride react to form 1 mol of lead chloride.

No. of moles of NaCl = m/M.M
= 5/58.5
= 0.0854

So, 0.0427 moles of PbCl2 are formed.

Mass of PbCl2 = n of PbCl2* M.M
= 0.0427 * 278.2
= 11.88g

Answer 3

The mass of lead chloride precipitated is 11.88 grams.

Answer 4

The balanced equation for this reaction is as follows:

2NaCl + Pb(NO3)2 ---> 2NaNO3 + Pb(Cl2)

So, 2 moles of sodium chloride react to form 1 mol of lead chloride.

No. of moles of NaCl = m/M.M
= 5/58.5
= 0.0854

So, 0.0427 moles of PbCl2 are formed.

Mass of PbCl2 = n of PbCl2* M.M
= 0.0427 * 278.2
= 11.88g

Answer 5

The mass of lead chloride precipitated is 11.88 grams.