Question

What mass of limestone (in kg) would be required to completely neutralize a 15.5 billion-liter lake that is 1.7×10^−5 M in H2SO4 and 8.9×10^−6 M in HNO3?

Answer 1

 

The complete balanced chemical reactions are:

HNO3 =>          CaCO3 + 2HNO3 → Ca(NO3)2 + H2O + CO2(g)

H2SO4 =>         CaCO3 + H2SO4 → CaSO4 + H2O + CO2(g)

 

So we see that 1 mole of CaCO3 is needed for 2 moles of HNO3 and similarly to 1 mole of H2SO4.
The number of moles can be calculated as the product of volume and molarity, so:

 

moles H2SO4 = 1.7×10^−5 M * (15.5 x 10^9 L) = 263,500 mol H2SO4

moles HNO3 = 8.9×10^−6 M * (15.5 x 10^9 L) = 137,950 mol HNO3

 

So the total moles of CaCO3 required is:

moles CaCO3 = 263,500 mol * 1 + 137,950 mol * (1/2)

moles CaCO3 = 332,475 mol

 

The molar mass of CaCO3 is 100.086 g/mol, so the mass is:

mass CaCO3 = 332,475 mol * 100.086 g/mol

mass CaCO3 = 33,276,092.85 g = 33.3 x 10^3 kg