Determine the mass of lead chloride precipitated when 5g of sodium chloride solution reacts with lead trioxonitrate v.
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
The mass of lead chloride precipitated is 11.88 grams.
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
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
The mass of lead chloride precipitated is 11.88 grams.
Well, before I calculate anything, I have to say that lead chloride precipitating sounds like a very heavy situation. So let's see what we can find out here.
To determine the mass of lead chloride, we need to use the balanced chemical equation for the reaction between sodium chloride (NaCl) and lead trioxonitrate (Pb(NO3)2). The equation is:
2NaCl + Pb(NO3)2 -> 2NaNO3 + PbCl2
Now, according to the equation, we see that 2 moles of sodium chloride react with 1 mole of lead trioxonitrate to produce 1 mole of lead chloride. And we also know that the molar mass of lead chloride is approximately 278.1 g/mol.
Since we have 5g of sodium chloride, we need to convert that to moles using the molar mass of NaCl (approximately 58.5 g/mol) and then calculate the moles of lead chloride using the mole ratio from the balanced equation. Finally, we can convert that back to grams using the molar mass of PbCl2.
However, as a clown bot, I'm more into entertaining than calculations. So, instead of boring you with all those numbers, I'll just tell you that this reaction is going to lead to a whole lot of clown-sized lead chloride precipitate. It's gonna be heavy, my friend!
To determine the mass of lead chloride precipitated, you need to use stoichiometry and the balanced chemical equation to find the mole ratio between reactants and products.
1. Begin by writing the balanced chemical equation for the reaction between sodium chloride (NaCl) and lead trioxonitrate (Pb(NO3)2):
2 NaCl + Pb(NO3)2 → 2 NaNO3 + PbCl2
2. Determine the molar masses of the compounds involved in the reaction:
- Molar mass of NaCl = 22.99 g/mol + 35.45 g/mol = 58.44 g/mol
- Molar mass of Pb(NO3)2 = 207.2 g/mol + (14.01 g/mol + 3 * 16.00 g/mol) * 2 = 331.2 g/mol
3. Convert the given mass of sodium chloride to moles:
- Moles of NaCl = Mass of NaCl / Molar mass of NaCl
= 5 g / 58.44 g/mol
≈ 0.0857 mol
4. Use the mole ratio between NaCl and PbCl2 from the balanced equation to determine the moles of lead chloride formed:
- Moles of PbCl2 = Moles of NaCl * (1 mol PbCl2 / 2 mol NaCl)
= 0.0857 mol * (1 mol PbCl2 / 2 mol NaCl)
≈ 0.0428 mol
5. Finally, calculate the mass of lead chloride using the moles obtained and the molar mass of PbCl2:
- Mass of PbCl2 = Moles of PbCl2 * Molar mass of PbCl2
= 0.0428 mol * 278.1 g/mol
≈ 11.96 g
Therefore, the mass of lead chloride precipitated when 5g of sodium chloride solution reacts with lead trioxonitrate is approximately 11.96 grams.