Question

Sodium carbonate (Na2CO3) is used to neutralize the sulfuric acid spill. How many kilograms of sodium carbonate must be added to neutralize 4.03×10³ kg of sulfuric acid solution?

Answer 1

H2SO4 + Na2CO3 -------------------> Na2SO4 + CO2 + H2O

1 mol of Na2CO3 is required ------------   to neutralise 1 mol of H2SO4 spilt.

mass of H2SO4 = 4.03 x 10^3 kg = 4.03 x 10^6 g

BUT it is only 95% H2SO4,

so mass H2SO4 = 95/100 x (4.03x10^6) g

= 3.8285x10^6 g

moles H2SO4 = mass / molar mass

= 3.829x10^6 g / 98.1 g/mol

= 3.903 x10^4 moles

So you will need 3.904 x 10^4 moles of Na2CO3 to neutralise it.

moles = mass / molar mass

mass of Na2CO3 = 3.903 x 10^4 moles x 106

= 4.14 x 10^6 g of Na2CO3

= 4.14 x 10^3 kg

mass of Na2CO3 = 4136.8 kg

= 4.14 x 10^3 kg