7.
HCl + NaOH --> H2O + NaCL
Molecular weight [MW]
MW NaOH = 40 g/mol
MW HCl= 36.45 g/mol
The steqiometry is 1 mol of HCl for 1 mol of NaOH, therefore:
moles NaOH = 0.040 [g] /40 [g/mol] = 0.001 mol
So we need 0.001 mol of HCl,
0.001(HCl) mol * 36.45 [g/mol] = 0.0364 g. So we need 0.0364 g of HCl to neutralize 0.04 g of NaOH
For 7 a) we need more information, we need the total amount of solvent (volume) so we can have the molarity.
7B)
H+(aq) + Cl-(aq) + Na+(aq) + OH-(aq) --> H2O(l) + Na+ (aq) + Cl-(aq)
We can only answer one question per exercise.
Help please How many grams of hydrochloric acid would it take to neutralize 0 040 grams...
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