Asked by Bella
When aqueous solutions of sodium sulfate and lead (II) nitrate are mixed lead sulfate precipitates out of soultion. Calculate the mass of lead (II) sulfate that should form when 1.25 L of 0.0500 M Pb(NO3)2 and 2.00 L of 0.0250 M Na2SO4 are reacted together.
Answers
Answered by
DrBob222
Pb(NO3)2+ Na2SO4 ==> PbSO4 + 2NaNO3
moles Pb(NO3)2 = 1.25L x 0.0500M = 0.0625.
moles Na2SO4 = 2L x 0.0250 = 0.0500.
Since 1 mol Pb(NO3)2 uses 1 mol Na2SO4 to form 1 mol PbSO4, mol PbSO4 formed must be 0.0500. There isn't enough Na2SO4 for all of the Pb(NO3)2 to be used.
Convert 0.0500 mols PbSO4 to grams. g= mols x molar mass = ?
moles Pb(NO3)2 = 1.25L x 0.0500M = 0.0625.
moles Na2SO4 = 2L x 0.0250 = 0.0500.
Since 1 mol Pb(NO3)2 uses 1 mol Na2SO4 to form 1 mol PbSO4, mol PbSO4 formed must be 0.0500. There isn't enough Na2SO4 for all of the Pb(NO3)2 to be used.
Convert 0.0500 mols PbSO4 to grams. g= mols x molar mass = ?
Answered by
Anonymous
15.2
Answered by
Tapanga
15.1633 g PbSO4
Answered by
gloria
15.2g
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