Asked by Ayo
Determine the mass of lead chloride precipitated when 5g of sodium chloride solution reacts with lead trioxonitrate v.
Answers
Answered by
Arora
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
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
Answered by
Olamide azeez
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
Answered by
Alabede Abdulazeez
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
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
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