Respuesta :
A reaction between acid and base to form water and salt is known as neutralization reaction. It is a double replacement reaction.
The reaction between [tex]H_2SO_4[/tex] and [tex]NaHCO_3[/tex] will be:
[tex]H_2SO_4 + NaHCO_3 \rightarrow Na_2SO_4 + CO_2 + H_2O[/tex]
The balanced reaction is:
[tex]H_2SO_4 + 2NaHCO_3 \rightarrow Na_2SO_4 + 2CO_2 + 2H_2O[/tex]
Volume of [tex]H_2SO_4[/tex] = 28 mL (given)
Since, 1 L = 1000 mL
So, 28 mL = 0.028 L
Molarity of [tex]H_2SO_4[/tex] = 5.4 M (given)
Molarity = [tex]\frac{number of moles of solute}{Volume of solution in Liters}[/tex] -(1)
Substituting the values in equation (1):
[tex]5.4 mol/L = \frac{number of moles of H_2SO_4}{0.028 L}[/tex]
[tex]number of moles of H_2SO_4 = 5.4 mol/L\times0.028 L[/tex]
[tex]number of moles of H_2SO_4 = 0.1512 mol[/tex]
From the balanced reaction between [tex]H_2SO_4[/tex] and [tex]NaHCO_3[/tex], 2 moles of [tex]NaHCO_3[/tex] reacts with 1 mole of [tex]H_2SO_4[/tex].
Molar mass of [tex]NaHCO_3[/tex] = 84.007 g/mol
Mass of [tex]NaHCO_3[/tex] needed:
[tex]0.1512 mol H_2SO_4 \times \frac{2 mole NaHCO_3}{1 mole H_2SO_4}\times \frac{84.007 g}{1 mole NaHCO_3}[/tex] = [tex]25.404 g[/tex]
Hence, the required amount of [tex]NaHCO_3[/tex] is [tex]25.404 g[/tex].
