Chemistry, 10.07.2019 11:40 amberpetty4288

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?

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Answer from: tyrique86

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

Answer from: jasmine2531

Mass of limestone required for complete neutralization is 3,336.64 $\rm \times\;10^4$ grams.

Limestone is $\rm CaCO_3$ . The neutralization reaction for limestone will be:

$\rm CaCO_3\;+\;2\;HNO_3\;\rightarrow\;Ca(NO_3)_2\;+\;H_2O\;+\;CO_2$

$\rm CaCO_3\;+\;H_2SO_4\;\rightarrow\;CaSO_4\;+\;H_2O\;+\;CO_2$

The balanced equation states that for neutralization of 1 mole of limestone, 2 moles $\rm HNO_3$ and 1 mole of $\rm H_2SO_4$ is required.

The moles of $\rm HNO_3$ and $\rm H_2SO_4$ available are:

Moles of $\rm HNO_3$ = molarity $\times$ volume (L)

Moles of $\rm HNO_3$ = 8.9 $\rm \times\;10^-^6$ $\times$ 15.5 $\rm \times\;10^9$ L

Moles of $\rm HNO_3$ = 13.975 $\rm \times\;10^4$ moles

Moles of $\rm H_2SO_4$ = 1.7 $\rm \times\;10^-^5$ $\times$ 15.5 $\rm \times\;10^9$ L

Moles of $\rm H_2SO_4$ = 26.35 $\rm \times\;10^4$ moles

Moles of $\rm CaCO_3$ required = 1 mole of $\rm H_2SO_4$ + $\rm \dfrac{1}{2}$ moles of $\rm HNO_3$

Moles of $\rm CaCO_3$ required = 26.35 $\rm \times\;10^4$ moles $\rm H_2SO_4$ + $\rm \dfrac{1}{2}$ (13.975 $\rm \times\;10^4$ moles) $\rm HNO_3$

Moles of $\rm CaCO_3$ required = 26.35 $\rm \times\;10^4$ + 6.9875 $\rm \times\;10^4$

Moles of $\rm CaCO_3$ required = 33.3375 $\rm \times\;10^4$ moles

Mass of limestone = moles of limestone $\times$ molecular weight of limestone

Mass of limestone = 33.3375 $\rm \times\;10^4$ moles $\times$ 100.0869 grams

Mass of limestone = 3,336.64 $\rm \times\;10^4$ grams.

For more information refer the link;

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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

Answer from: Quest

the correct answer is option d

explanation:

i was wrong its d because they dont use cm when calculating square roots

40 points and brainliest to anyone who can answer these

Answer from: Quest

a saline water is fed to a reverse osmosis unit at flow rate o rotameter connected to reject (retentate) line reads 8 installed in reject and permeate line reads 15.2 milli siemens/cm and 300 micro siemens/com respectively.

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What mass of limestone (in kg) would be required to completely neutralize a 15.5 billion-liter lake...