Question

A tanker truck carrying 6.05×103 kg of concentrated sulfuric acid solution tips over and spills its load.
The sulfuric acid solution is 95.0% H2SO4    by mass and has a density of 1.84 g/mL .
Sodium carbonate (Na2O3 ) is used to neutralize the sulfuric acid spill. How many kilograms of sodium carbonate must be added to neutralize 6.05×103 kg of sulfuric acid solution?

Answer 1

Given that;

6.05×10^3 kg of concentrated sulfuric acid

The sulfuric acid solution is 95.0% H2SO4    by mass and has a density of 1.84 g/mL .

First calculate the mass of H2SO4 as follows:

6.05×10^3 kg = 6.05×10^6 g

1.00 kg = 1000 g

m(H2SO4) = %H2SO4 x m(solution)
m(H2SO4) = 0.95 x 6.05×10^6 g

m(H2SO4) = 5.748x10^6 g of H2SO4

Now calculate the number of mole of H2SO4:

Number of moles = amount in g / molar mass

=5.748x10^6 g of H2SO4 / 98.078 g/ mole

= 58601 mol of H2SO4

the balance reaction between H2SO4 and Na2CO3 is as follows:

H2SO4 + Na2CO3 = Na2SO4 + CO2 + H2O


then calculate the mole of Na2CO3:

= 58601 mol of H2SO4 *1/1

= 58601 mol of Na2CO3

Amount in g = number of moles * molar mass

=58601 mol of Na2CO3 *105.99 g/ mole

= 621115.623 g

= 6211.15 kg

= 6.21*10^3 kg of Na2CO3