0.63g of pb powder were disolved in excces nitrate(v) aci to forn leadnitrate solution, all the lead nitrate solution were reacted with sulphate,give the ion equationof the reaction between lead nitrate and sadium sulphate solution?determine the mass of the lead salt formed in above (pb207,s42,o16)?
You need to find the caps key on your keyboard AND USE IT. I don't get the s42 stuff. And when you say lead salt in the above you COULD mean Pb(NO3)2 or PbSO4. I will assume you mean PbSO4.
Pb^2+ + SO4^2- --> PbSO4.
mols Pb^2+ = grams/molar mass = ?
Since the ratio is 1 Pb^2+ to 1 PbSO4 formed, then mols PbSO4 = Pb^2+.
Then grams PbSO4 = mols PbSO4 x molar mass PbSO4 = ?
By the way, you need to spell chemistry right also. Further, I don't think this is the first time I've reminded you of that.
To determine the ion equation of the reaction between lead nitrate and sodium sulfate, we need to know the chemical formulas for both compounds involved.
Lead nitrate is composed of the ions Pb²⁺ (lead) and NO₃⁻ (nitrate). Sodium sulfate consists of the ions Na⁺ (sodium) and SO₄²⁻ (sulfate).
The balanced ionic equation for the reaction can be determined by swapping the anions and balancing the charges:
Pb²⁺ + SO₄²⁻ → PbSO₄
Therefore, the ion equation for the reaction between lead nitrate and sodium sulfate is Pb²⁺ + SO₄²⁻ → PbSO₄.
To determine the mass of the lead salt formed, we need to use the molar masses of the compounds involved.
The molar mass of PbSO₄ is calculated by adding up the molar masses of its constituent elements. The molar mass of lead (Pb) is 207 g/mol, sulfur (S) is 32 g/mol, and oxygen (O) is 16 g/mol.
Molar mass of PbSO₄ = (207 g/mol) + (32 g/mol) + (4 × 16 g/mol) = 303 g/mol
To calculate the mass of the lead salt formed, we can use the given information that 0.63 g of Pb powder were dissolved in excess nitrate(V) acid. Since the molar mass of Pb powder is not provided, we assume it to be the same as the molar mass of Pb.
Using the formula: Mass = Moles × Molar mass
Moles of Pb powder = Mass of Pb powder / Molar mass of Pb
Moles of Pb powder = 0.63 g / 207 g/mol ≈ 0.00304 mol
Since the balanced equation shows a 1:1 ratio between Pb powder and PbSO₄, the moles of PbSO₄ formed will be the same as the moles of Pb powder.
Mass of PbSO₄ = Moles of PbSO₄ × Molar mass of PbSO₄
Mass of PbSO₄ = 0.00304 mol × 303 g/mol ≈ 0.923 g
Therefore, the mass of the lead salt (PbSO₄) formed in the reaction is approximately 0.923 g.