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?

Answers: 1

Show answers

Answers

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;

link

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

by the combination of an unlimited number of elements

Answer from: Quest

2b2t

explanation:

2b2t

Another question on Chemistry

Chemistry, 22.06.2019 06:00

In 1901, thomas edison invented the nickel-iron battery. the following reaction takes place in the battery. fe(s) + 2 nio(oh)(s) + 2 h2o(l) fe(oh)2(s) + 2 ni(oh)2(aq) how many mole of fe(oh)2, is produced when 5.35 mol fe and 7.65 mol nio(oh) react?

Answers: 3

Answer

Chemistry, 22.06.2019 08:00

Why are pipes bursting in the in extremely cold weather?

Answers: 2

Answer

Chemistry, 22.06.2019 13:00

Which of the following are good traits of a hypothesis? it will be able to be testedit can predict an outcomeit will explain the observationsall of these

Answers: 2

Answer

Chemistry, 22.06.2019 23:30

If it is an isoelectronic series select true, if not select false. o2-, s2-, se2-, te2- na+, k+, rb+, cs+ n3-, p3-, as3-, sb3- ag, cd+, sn3+, sb4+ f-, cl-, br-, i- f-, ne, na+, mg2+ s2-, s, s6+

Answers: 1

Answer

You know the right answer?

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

Questions