Question

According to the label on a bottle of concentrated hydrochloric acid, the contents are 36% by mass and have a density of 1.138g/mL. What mass (in Kg) of sodium hydrogen carbonate would be needed to neutralize the spill if a bottle containing 1.508L of concentrated HCl dropped on the lab floor and broke open?

Answer 1

1.42 Kg of NaHCO3

Step-by-step explanation:

From

Co= 10 pd/M

Where;

Co= concentration of the stock solution of acid

p= percentage of raw acid = 36%

d= density of the acid= 1.138g/mL

M= molar mass of the acid= 36.5 g/mol

Co= 10 × 36 × 1.138/36.5

Co= 11.22 M

Number of moles of acid= CV = 11.22 M × 1.508 L= 16.9 moles

The reaction equation is;

NaHCO3(aq)+HCl(aq)→NaCl(aq)+H2O(l)+CO2(g)

If 1 mole of NaHCO3 reacts with 1 mole of HCl

Then 16.9 moles of HCl reacts with 16.9 moles of NaHCO3

Hence 16.9 moles of NaHCO3 is required to neutralize the HCl.

From

n= m/ M

Where;

n= number of moles = 16.9

m= mass of NaHCO3= ??

M= molar mass of NaHCO3= 84.007 g/mol

m= n× M/1000

m= 16.9 × 84.007 /1000

m= 1.42 Kg