Question

If 27 ml of 6.0 m h2so4 was spilled, what is the minimum mass of nahco3 that must be added to the spill to neutralize the acid?

Answer 1

The minimum mass of NaHCO3 that  must be added to the spill to neutralize the acid is 27.216 grams

calculation

write the balanced chemical equation

2NaHCO3 +H2SO4 → Na2SO4  +2H2O +2CO2

find the moles of H2SO4 = molarity x volume in liters

volume in liters =  27/1000=0.027 l

moles is therefore= 0.027 x6=0.162 moles

by use of mole ratio of  NaHCO3: H2SO4  which is 2:1 the moles of NaHCO3=0.162 x2=0.324 moles

mass of NaHCO3= moles of NaHCO3 x molar mass of NaHCO3(84g/mol)

= 84g/mol x 0.324=27.216 grams

Answer 2

The formula of molarity is:

$molarity = \frac{number of moles of solute}{Volume of solvent in L}$   -(1)

Molarity of $H_2SO_4 = 6.0 M$     (given)

Volume of $H_2SO_4 = 27.0 mL$     (given)

Since, $1 L = 1000 mL$

So, $27 mL = 0.027 L$

Substituting the values in formula (1):

$6.0 M = \frac{number of moles of H_2SO_4}{0.027 L}$

$number of moles of H_2SO_4 = 0.027 L\times 6.0 mol/L = 0.162 mol$

The balanced chemical reaction between $H_2SO_4$ and $NaHCO_3$ is:

$H_2SO_4+2NaHCO_3\rightarrow Na_2SO_4+2CO_2+2H_2O$

From the above reaction it is clear that, 2 moles of $NaHCO_3$ reacts with 1 mole of $H_2SO_4$. So, for $0.162 mol$ of $H_2SO_4$ number of moles of $NaHCO_3$ is:

Number of moles of $NaHCO_3$ = $0.162 \times 2 = 0.324 mol$

Since, $Moles = \frac{mass}{Molar Mass}$   -(2)

Molar mass of $NaHCO_3$ = $23 + 1+12+(3\times 16) = 84 g/mol$

Using formula (2):

Mass of $NaHCO_3$ = Moles of $NaHCO_3\times$ Molar mass of $NaHCO_3$

Substituting the values:

Mass of $NaHCO_3$ = $0.324 mol\times 84g/mol = 27.216 g$

Hence, $27.216 g$ of $NaHCO_3$ must be added to neutralize the spill acid.