Question

A site in Pennsylvania receives a total annual deposition of 2.688 g/mof sulfate from fertilizer and acid rain. The ratio by mass of ammonium sulfate/ammonium bisulfate/sulfuric acid is 3.0/5.5/1.0. (b) How many pounds of CaCO₃ are needed to neutralize this acid?

Answer 1

According to the statement

2.12 x 10^4 lbs pounds of CaCO₃ are needed to neutralize this acid

What is neutralization?

A chemical reaction in which an acid and a base react quantitatively with each other is known as neutralization or neutralization. In a water reaction, neutralization ensures that there is no excess of hydrogen or hydroxide ions in the solution.

According to the given information:

The equation of the neutralization reaction between H2SO4 and CaCO3.

CaCO3 + H2SO4 → CaSO4 + H2CO3

H2CO3 dissociate to water and carbon dioxide.

        CaCO3 + H2SO4 → CaSO4  + H2O + CO2

Now solving for the mass of CaCO3 needed to neutralize the acid.

mass of CaCO3 = 9460 Kg H2SO4  × $\frac{1000 \mathrm{~g}}{1 \mathrm{~kg}} \times \frac{1 \mathrm{~mol} \mathrm{H}_2 \mathrm{SO} 4}{98.1 \mathrm{gH}_2 \mathrm{SO}_4} \times \frac{1 \mathrm{~mol} \mathrm{CaCO}\left(\mathrm{O}_3\right.}{1 \mathrm{~mol} \mathrm{H}_2 \mathrm{SO}_4}$$\times \frac{100.1 \mathrm{~g} \mathrm{CaCO}_3}{1 \mathrm{~mol} \mathrm{CaCO}_3} \times \frac{2.205 \mathrm{lb}}{1000 \mathrm{~g}}$

= 21284.56606

mass of CaCO3 =  2.12 x 10^4 lbs

2.12 x 10^4 lbs pounds of CaCO₃ are needed to neutralize this acid.

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