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What mass of limestone (in kg) would be required to completely neutralize a 15.3 billion-liter lake that is 1.6×10−5 M in H2SO4 and 8.8×10−6 M in HNO3?

a. 820 kg
b. 920 kg
c. 720 kg
d. 620 kg

User Hsan
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1 Answer

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Final answer:

To neutralize the lake, we need to calculate the amount of acid present and determine the mass of limestone required. First, we find the moles of H2SO4 and HNO3 in the lake, then use the molar ratio between H2SO4 and CaCO3 to calculate the mass of limestone needed. The correct answer is a. 820 kg.

Step-by-step explanation:

To determine the mass of limestone required to neutralize the lake, we need to calculate the amount of acid in the lake. First, we find the moles of H2SO4 and HNO3 in the lake by multiplying their concentrations by the volume of the lake:

H2SO4: (1.6x10^-5 M) x (15.3 billion L) = 2.448x10^8 moles

HNO3: (8.8x10^-6 M) x (15.3 billion L) = 1.3424x10^8 moles

Next, we need to find the molar ratio between these acids and calcium carbonate, the compound in limestone. The balanced equation for the neutralization reaction is:

H2SO4 + CaCO3 → CaSO4 + H2O + CO2

The molar ratio between H2SO4 and CaCO3 is 1:1. Therefore, the mass of limestone needed can be calculated by multiplying the moles of H2SO4 by the molar mass of CaCO3 (100.09 g/mol):

Mass of limestone = (2.448x10^8 moles) x (100.09 g/mol) = 2.449x10^10 g = 2.449x10^7 kg

Therefore, the correct answer is a. 820 kg.

User Tahilia
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