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?

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

Mass of limestone required for complete neutralization is 3,336.64 grams.

Limestone is . The neutralization reaction for limestone will be:

The balanced equation states that for neutralization of 1 mole of limestone, 2 moles and 1 mole of is required.

The moles of and available are:

Moles of = molarity volume (L)

Moles of = 8.9 15.5 L

Moles of = 13.975 moles

Moles of = 1.7   15.5 L

Moles of = 26.35   moles

Moles of required = 1 mole of + moles of

Moles of required = 26.35   moles + (13.975 moles)

Moles of required = 26.35   + 6.9875  

Moles of required = 33.3375   moles

Mass of limestone = moles of limestone molecular weight of limestone

Mass of limestone =  33.3375   moles 100.0869 grams

Mass of limestone = 3,336.64 grams.

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 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

the atomic mass is the bottom number on the element.

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the answer to your question is c