Question
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)?
Answers
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 = ?
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.
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