Asked by Walker
A solution of 100.00mL of 0.2000mol/L sodium carbonate and 200.00mL of 0.1000mol/L calcium nitrate solutions are mixed together according to the reaction: sodium carbonate + calcium nitrate→ calcium carbonate + sodium nitrate. how would you calculate the mass of calcium carbonate that would precipitate?
Answers
Answered by
DrBob222
A limiting reagent problem.
Na2CO3 + Ca(NO3)2 ==> CaCO3 + 2NaNO3
mol Na2CO3 = M x L = 0.200 x 0.100 = 0.02
mol Ca(NO3)2 = M x L = 0.100 x 0.200 = 0.02 mol.
Calculate how much CaCO3 is formed if we use 0.02 mol Na2CO3 and all the Ca(NO3)2 needed. That will be
0.02 mol Na2CO3 x (1 mol CaCO3/1 mol Na2CO3) = 0.02 mol CaCO3 formed.
Next calculate how much CaCO3 is formed if we use 0.02 mol Ca(NO3)2 and all the Na2CO3 needed. That will be
0.02 mol Ca(NO3)2 x (1 mol CaCO3/1 mol Ca(NO3)2) = 0.02 mol.
Since the number is the same (0.02 mol CaCO3), neither Ca(NO3)2 nor Na2CO3 is the limiting reagent and 0.02 mol CaCO3 is formed.
g CaCO3 = mols x molar mass = ?
Na2CO3 + Ca(NO3)2 ==> CaCO3 + 2NaNO3
mol Na2CO3 = M x L = 0.200 x 0.100 = 0.02
mol Ca(NO3)2 = M x L = 0.100 x 0.200 = 0.02 mol.
Calculate how much CaCO3 is formed if we use 0.02 mol Na2CO3 and all the Ca(NO3)2 needed. That will be
0.02 mol Na2CO3 x (1 mol CaCO3/1 mol Na2CO3) = 0.02 mol CaCO3 formed.
Next calculate how much CaCO3 is formed if we use 0.02 mol Ca(NO3)2 and all the Na2CO3 needed. That will be
0.02 mol Ca(NO3)2 x (1 mol CaCO3/1 mol Ca(NO3)2) = 0.02 mol.
Since the number is the same (0.02 mol CaCO3), neither Ca(NO3)2 nor Na2CO3 is the limiting reagent and 0.02 mol CaCO3 is formed.
g CaCO3 = mols x molar mass = ?
Answered by
Ruth.
1.201g
Answered by
David
This answer seems to be correct but you should be explaining in details please
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