Asked by Lizzy
A 150-mL sample of a .600-mol/L calcium nitrate solution is mixed with a 1.00L of .500-mol/L NaOH(aq). Calculate the mass of precipitate produced in this reacton.
Answers
Answered by
DrBob222
mols Ca(NO3)2 = M x L = ?
mols NaOH = M x L = ?
.....Ca(NO3)2 + 2NaOH ==> Ca(OH)2 + 2H2O
Convert mols Ca(NO3)2 to mols Ca(OH)2.
Convert mols NaOH to mols Ca(OH)2.
In limiting reagent problems the smaller amount is ALWAYS the correct amount of ppt formed.
g ppt formed = mols x molar mass.
mols NaOH = M x L = ?
.....Ca(NO3)2 + 2NaOH ==> Ca(OH)2 + 2H2O
Convert mols Ca(NO3)2 to mols Ca(OH)2.
Convert mols NaOH to mols Ca(OH)2.
In limiting reagent problems the smaller amount is ALWAYS the correct amount of ppt formed.
g ppt formed = mols x molar mass.
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