No, the reactants acetic acid (C2H4O2) and sodium bicarbonate (NaHCO3) react chemically to make other compounds (products): sodium acetate (NaC2H3O2) and carbonic acid (H2CO3), decomposing easily to carbon dioxide (CO2) and water (H2O)
NaHCO3 is sodium bicarbonate (baking soda.) It is water soluble, and in solution it will dissociate into Na+ ions and HCO3- ions.
Antacids contain carbonates as Na2CO3 and NaHCO3.
60.04
Several part problem. Get molarity of NaHCO3. (150 ml)( M NaHCO3) = (150 ml)(0.44 M HCl) = 0.44 M NaHCO3 --------------------------- get moles NaHCO3 ( 150 ml = 0.150 Liters ) 0.44 M NaHCO3 = moles NaHCO3/0.150 Liters = 0.066 moles NaHCO3 ---------------------------------------get grams 0.066 moles NaHCO3 (84.008 grams/1 mole NaHCO3) = 5.54 grams NaHCO3 needed ---------------------------------------------answer
Physical
NaHCO2 is an unknown compound, NaHCO3 is sodium bicarbonate (baking powder). The answer is given for the 'corrected' question.
No, the reactants acetic acid (C2H4O2) and sodium bicarbonate (NaHCO3) react chemically to make other compounds (products): sodium acetate (NaC2H3O2) and carbonic acid (H2CO3), decomposing easily to carbon dioxide (CO2) and water (H2O)
It is a chemical change. CH3COOH (vinegar) + NaHCO3 (baking soda) -> CH3COONA (sodium acetate) + H2CO3 (carbonic acid)...which then immediately dissociates to... H2CO3 -> H2O + CO2
NaHCO3 is sodium bicarbonate (baking soda.) It is water soluble, and in solution it will dissociate into Na+ ions and HCO3- ions.
Any change (NaHCO3)
As an overview, acetic acid (HC2H3O2 (or C2H4O2 in solution), from the vinegar) reacts with aqueous sodium bicarbonate (NaHCO3), forming sodium acetate (NaC2H3O2) and carbonic acid (H2CO3). As also happens in carbonated water, the carbonic acid then dissociates into water and carbon dioxide (H2O and CO2). The chemical reaction would be (not balanced): C2H4O2 + NaHCO3 --initial reaction--> NaC2H3O2 + H2CO3 --bubbling--> NaC2H3O2 + H2O + CO2
5.00 moles H x 1 mole C2H4O2/4 moles H = 1.25 moles of C2H4O2 present.
organic
Antacids contain carbonates as Na2CO3 and NaHCO3.
60.04
Several part problem. Get molarity of NaHCO3. (150 ml)( M NaHCO3) = (150 ml)(0.44 M HCl) = 0.44 M NaHCO3 --------------------------- get moles NaHCO3 ( 150 ml = 0.150 Liters ) 0.44 M NaHCO3 = moles NaHCO3/0.150 Liters = 0.066 moles NaHCO3 ---------------------------------------get grams 0.066 moles NaHCO3 (84.008 grams/1 mole NaHCO3) = 5.54 grams NaHCO3 needed ---------------------------------------------answer
It is a Physical change