I have worked this over and over the answer is 37 g. but I am not coming up with that can anyone help?
A flash containing 450 ml of 0.50 M H2SO4 Was knocked onto floor. How many grams of NaHCO3 do you need to put on the spill to neutralize the acid according to the following equation?
H2SO4 (aq) + 2 NaHCO3(aq) to Na3Ag(S2O3)2(aq + NaBr (aq)
The equation does not look correct.
H2SO4 (aq) + 2 NaHCO3(s) -> Na2SO4(aq) + 2H2O(l) + 2CO2(g)
the spill contains
450 x 0.50 mmoles of H2SO4 = 225 mmole of H2SO4
we 2 moles of NaHCO3 for each mole of H2SO4 from the equation.
so 2 x 225 mmoles needed = 450 mmoles or 0.450 mole
molar mass of NaHCO3 is 84 g mol^-1
so mass needed =0.450 mol x 84 g mol^-1
= 37.8 g