What mass of lead (ii)trioxonitrate(v)pb (No3)2 would be required to yield 25g of pbcl on the addition of excess of Nacl solution (Pb=207,N=14,O=16,Na=23,Cl=35.5)
If your equation is
Pb(NO3)2 + 2NaCl → PbCl + 2NaNO3
then each mole of Pb(NO3)2 yields one mole of PbCl.
So, how many moles of PbCl do you have in 25 grams?
convert that to grams of Pb(NO3)2
You need to start over. Lead(II) chloride is PbCl2. Enough said?
And lead (II) trioxonitrate(V) is not an IUPAC approved name for lead(II) nitrate
To determine the mass of lead(II) trioxonitrate(V) Pb(NO3)2 required to yield 25g of PbCl2, we need to set up a balanced chemical equation for the reaction between lead(II) trioxonitrate(V) and sodium chloride (NaCl). The balanced equation for this reaction is as follows:
2Pb(NO3)2 + 2NaCl → 2PbCl2 + 4NaNO3
From this equation, we can see that 2 moles of lead(II) trioxonitrate(V) Pb(NO3)2 react with 2 moles of sodium chloride (NaCl) to produce 2 moles of lead chloride (PbCl2) and 4 moles of sodium nitrate (NaNO3).
Now, let's calculate the molar masses of the compounds involved:
Molar mass of Pb(NO3)2:
Pb = 207g/mol
N = 14g/mol (x2 for subscript)
O = 16g/mol (x6 for subscript)
Total molar mass = 207 + (14x2) + (16x6) = 207 + 28 + 96 = 331g/mol
Molar mass of NaCl:
Na = 23g/mol
Cl = 35.5g/mol
Total molar mass = 23 + 35.5 = 58.5g/mol
Now, let's calculate the number of moles of PbCl2 required to yield 25g:
Molar mass of PbCl2:
Pb = 207g/mol
Cl = 35.5g/mol (x2 for subscript)
Total molar mass = 207 + (35.5x2) = 207 + 71 = 278g/mol
Number of moles of PbCl2 = mass / molar mass = 25g / 278g/mol ≈ 0.090 moles
From the balanced equation, we can see that 2 moles of Pb(NO3)2 yield 2 moles of PbCl2. Therefore, the number of moles of Pb(NO3)2 required is also 0.090 moles.
Finally, let's calculate the mass of Pb(NO3)2 required:
Mass = number of moles * molar mass = 0.090 moles * 331g/mol ≈ 29.79g
Therefore, approximately 29.79 grams of lead(II) trioxonitrate(V) Pb(NO3)2 would be required to yield 25g of PbCl2 when an excess of NaCl solution is added.