Question

Lakes that have been acidified by acid rain can be neutralized by the addition oflimestone (CaCO3). How much limestone in kg would be required to completelyneutralize a 5.2 x 10^9 L lake containing 5.0 x 10-3 g of H2SO4 per liter?

Answer 1

2.651 * 10⁴kg

Explanations:

The balanced reaction between limestone (CaCO₃) and sulfuric acid (H₂SO₄) is given as:

`CaCO_3+H_2SO_4\rightarrow H_2CO_3+CaSO_4`

Given the following parameters

• volume of sulfuric acid = 5.2 x 10⁹ L

,

• Mas of sulfuric acid = 0.005g

Determine the moles of sulfuric acid.

`\begin{gathered} moles\text{ of H}_2SO_4=5.2*10^9L*(0.005g)/(L)*(mol)/(98.079) \\ moles\text{ of }H_2SO_4=2.651*10^5moles \end{gathered}`

According to stochiometric, 1 mole of limestone reacted with 1mole of sulfuric acid, hence the number of moles of limestone required is 2.651 * 10⁵moles

Determine the required mass of limestone

`\begin{gathered} Mass\text{ of }CaCO_3=moles* molar\text{ mass} \\ Mass\text{ of C}aCO_3=2.651*10^5g*(100.09g)/(mol)\frac{}{} \\ Mass\text{ of CaCO}_3=2.651*10^7grams \end{gathered}`

Convert the result to kilogram

Recall that 1000g =1kg

Hence;

`\begin{gathered} Mass\text{ of CaCO}_3=2.651*(10^7)/(10^3)kg \\ Mass\text{ of CaCO}_3=2.651*10^4kg \end{gathered}`

Hence the mass of limestone in kg would be required to completely neutralize a 5.2 x 10⁹ L lake containing 5.0 x 10-3 g of H₂SO₄ per liter is 2.651 * 10⁴kg