Question

Solving for row three. To answer for column b we Are assuming that the concentration of acidic acid was 0.42 moles and the volume of acetic acid was 0.050 L

Answer 1

So, solving for row three, we are given that:

The mass of NaHCO3 is 1.5g.

To convert this amount to moles, we divide by the molecular mass of NaHCO3, which is 84g/mol.

So,

$molesNaHCO_3=\frac{1.5g}{\frac{84g}{\text{mol}}}=1.78\cdot10^{-2}$

And, given that the moles of C2H3O2H are:

$molesC_2H_3O_2H=2.1\cdot10^{-2}$

The amount of moles of NaHCO3 is less than the amount of moles of C2H3O2H, so, the limiting reagent is NaHCO3 and the excess reagent is C2H3O2H.