Asked by Joel
How do I calculate the mass of sodium carbonate required to neutralize 15-20 mL
of 0.25M HCl at the stoichiometric point?
of 0.25M HCl at the stoichiometric point?
Answers
Answered by
DrBob222
Na2CO2 + 2HCl ==> 2NaCl + CO2 + H2O
Say 20 mL. How many mols is that? M x L = mols HCl.
mols Na2CO3 = 1/2 that.
g Na2CO3 = mols x molar mass.
Say 20 mL. How many mols is that? M x L = mols HCl.
mols Na2CO3 = 1/2 that.
g Na2CO3 = mols x molar mass.
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