Solution :-
Balanced reaction equation is as follows
NaHCO3(s) + CH3COOH(aq) ------------ > NaCH3CO2(aq) + H2O(l) + CO2(g)
maximum volume = 4.5 L
and effective inflation percent is 95 %
therefore 4.5 L * 95 % /100 % = 4.275 L
now lets calculate the moles of CO2 at STP in 4.275 L
at STP 1 mol = 22.4 L
therefore
4.275 L * 1 mol / 22.4 L = 0.1908 mol CO2
now lets find the amount of each reactant needed
mole ratio of the CO2 to acetic acid and NaHCO3 is 1 : 1
therefore moles of the CH3COOH needed = 0.1908 mol
moles of NaHCO3 needed = 0.1908 mol
since acetic acid is 0.250 molar
therefore volume of acetic acid needed is
volume of CH3COOH = 0.1908 mol / 0.250 mol per L = 0.763 L
mass of NaHCO3 needed = moles * molar mass
= 0.1908 mol * 84.007 g per mol
= 16.0 g NaHCO3
There we need 0.763 L of 0.250 molar acetic acid (CH3COOH)
and 16.0 g of NaHCO3
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