2.651 * 10⁴kg
The balanced reaction between limestone (CaCO₃) and sulfuric acid (H₂SO₄) is given as:
[tex]CaCO_3+H_2SO_4\rightarrow H_2CO_3+CaSO_4[/tex]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.
[tex]\begin{gathered} moles\text{ of H}_2SO_4=5.2\times10^9L\times\frac{0.005g}{L}\times\frac{mol}{98.079} \\ moles\text{ of }H_2SO_4=2.651\times10^5moles \end{gathered}[/tex]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
[tex]\begin{gathered} Mass\text{ of }CaCO_3=moles\times molar\text{ mass} \\ Mass\text{ of C}aCO_3=2.651\times10^5g\times\frac{100.09g}{mol}\frac{}{} \\ Mass\text{ of CaCO}_3=2.651\times10^7grams \end{gathered}[/tex]Convert the result to kilogram
Recall that 1000g =1kg
Hence;
[tex]\begin{gathered} Mass\text{ of CaCO}_3=2.651\times\frac{10^7}{10^3}kg \\ Mass\text{ of CaCO}_3=2.651\times10^4kg \end{gathered}[/tex]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
