what mass of lead || trioxonitrate(v) would be required to yield 12g of lead || chloride on the addition of excess sodium chloride solution (pb=207,N=1,o=16,Na=23,cl=35.5) pb(No)+2Nacl pbcl+2NaNo
That chemical with the funny name is properly called lead(II) nitrate.
Pb(NO3)2 + 2NaCl ==> PbCl2 + 2NaNO3.
mols PbCl2 needed = grams/molar mass = ?
You get 1 mol PbCl2 for every 1 mol Pb(NO3)2 used initially.
Then grams Pb(NO3)2 = mols Pb(NO3)2 x molar mass Pb(NO3)2 = ?
Post your work if you get stuck.
did the solution
What is the mass
To determine the mass of lead(II) trioxonitrate(V) required to yield 12g of lead(II) chloride, we first need to balance the chemical equation:
Pb(NO3)2 + 2NaCl → PbCl2 + 2NaNO3
From the balanced equation, we can see that 1 mole of lead(II) trioxonitrate(V) reacts with 1 mole of lead(II) chloride. Therefore, we need to find the molar mass of lead(II) chloride to convert the given mass into moles.
The molar mass of PbCl2 is calculated as follows:
- The atomic mass of Pb = 207g/mol
- The atomic mass of Cl = 35.5g/mol (since there are two chlorine atoms, we multiply by 2)
- Total molar mass of PbCl2 = 207g/mol + 2*(35.5g/mol) = 207g/mol + 71g/mol = 278g/mol
To find the number of moles of lead(II) chloride:
Number of moles = mass / molar mass
Number of moles = 12g / 278g/mol ≈ 0.043mol
Since 1 mole of lead(II) trioxonitrate(V) reacts with 1 mole of lead(II) chloride, we know that 0.043 moles of lead(II) trioxonitrate(V) are required. Now we can calculate the mass of lead(II) trioxonitrate(V).
The molar mass of Pb(NO3)2 is calculated as follows:
- The atomic mass of Pb = 207g/mol
- The atomic mass of N = 14g/mol
- The atomic mass of O = 16g/mol (since there are three oxygen atoms, we multiply by 3)
- Total molar mass of Pb(NO3)2 = 207g/mol + 14g/mol + 3*(16g/mol) = 207g/mol + 14g/mol + 48g/mol = 269g/mol
To find the mass of lead(II) trioxonitrate(V):
Mass = number of moles * molar mass
Mass = 0.043mol * 269g/mol ≈ 11.57g
Therefore, approximately 11.57g of lead(II) trioxonitrate(V) would be needed to yield 12g of lead(II) chloride when excess sodium chloride solution is added.