Question

What mass of NaHCO3 would be needed to neutralize a spill consisting of 25.2mL of 6.01 M hydrochloric acid solution?

Answer 1

Approx. `13*g`....

Explanation:

We address the stoichiometric equation....

`Na^+HCO_3^(-) +HCl(aq)rarrNaCl(aq) +H_2O(l) + CO_2(g)uarr`

And thus there is 1:1 molar equivalence between hydrochloric acid and sodium bicarbonate...

`n_"HCl"=25.2*mLxx10^-3*L*mL^-1xx6.01*mol*L^-1=0.152*mol`

`"mass"_"sodium bicarbonate"=0.152*molxx84.01*g*mol^-1=??*g`