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Ainat [17]
3 years ago
6

If the pH of a 1.00-in. rainfall over 1800 miles2 is 3.70, how many kilograms of sulfuric acid, H2SO4, are present, assuming tha

t it is the only acid contributing to the pH? For sulfuric acid, Ka1 is very large and Ka2 is 0.012.
Chemistry
2 answers:
ArbitrLikvidat [17]3 years ago
7 0

<u>Ans: Mass of H2SO4 present = 231.5*10⁴ kg</u>

Height of rainfall = 1.00 inch = 0.0254 m

Area of rainfall = 1800 miles²

1 mile² = 2.59*10⁶ m²

Therefore area = 1800*2.59*10⁶ = 4.662*10⁹ m²

Volume (V) of rainfall = area (A) * height(h) = 0.0254 * 4.662*10⁹ = 1.184*10⁸ m³

1 m³ = 1000 L

V = 1.184*10¹¹ L

It is given that the pH = 3.70

pH = -log[H+]

[H+] = 10^-pH = 10⁻³.⁷⁰ = 1.995*10⁻⁴M

Now:

Molarity of H2SO4 = moles of H2SO4/volume

moles = 1.995*10⁻⁴moles.L-1 * 1.184*10¹¹ L = 2.362*10⁷ moles

Molar mass of H2SO4 = 98 g/mol

Mass of H2SO4 = 2.362*10⁷ moles * 98 g/mol = 231.48*10⁷ g = 231.5 *10⁴ kg

NARA [144]3 years ago
3 0
There are 2.32 x 10^6 kg sulfuric acid in the rainfall. 

Solution: 
We can find the volume of the solution by the product of 1.00 in and 1800 miles2: 
     1800 miles2 * 2.59e+6 sq m / 1 sq mi = 4.662 x 10^9 sq m 
     1.00 in * 1 m / 39.3701 in = 0.0254 m  
     Volume = 4.662 x 10^9 m^2 * 0.0254 m
                  = 1.184 x 10^8 m^3 * 1000 L / 1 m3
                  = 1.184 x 10^11 Liters 

We get the molarity of H2SO4 from the concentration of [H+] given by pH = 3.70: 
     [H+] = 10^-pH = 10^-3.7 = 0.000200 M 
     [H2SO4] = 0.000100 M  
 
By multiplying the molarity of sulfuric acid by the volume of the solution, we can get the number of moles of sulfuric acid: 
     1.184 x 10^11 L * 0.000100 mol/L H2SO4 = 2.36 x 10^7 moles H2SO4 

We can now calculate for the mass of sulfuric acid in the rainfall: 
     mass of H2SO4 = 2.36 x 10^7 moles * 98.079 g/mol
                               = 2.32 x 10^9 g * 1 kg / 1000 g
                               = 2.32 x 10^6 kg H2SO4
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The pH of the solution is 4.60.

Explanation:

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2 years ago
This question involves two calculations. The answer to the first part will be
ozzi

Answer :

(a) The mass of Al_2O_3 produced is, 15.2 grams.

(b) The percent yield of the reaction is, 72.5 %

Explanation :

Part (a) :

Given,

Mass of Al = 85.1 g

Molar mass of Al = 27 g/mol

First we have to calculate the moles of Al

\text{Moles of }Al=\frac{\text{Given mass }Al}{\text{Molar mass }Al}=\frac{85.1g}{27g/mol}=3.15mol

Now we have to calculate the moles of Al_2O_3

The balanced chemical equation is:

4Al+3O_2\rightarrow 2Al_2O_3

From the reaction, we conclude that

As, 4 moles of Al react to give 2 moles of Al_2O_3

So, 3.15 moles of Al react to give \frac{2}{4}\times 3.15=1.58 mole of Al_2O_3

Now we have to calculate the mass of Al_2O_3

\text{ Mass of }Al_2O_3=\text{ Moles of }Al_2O_3\times \text{ Molar mass of }Al_2O_3

Molar mass of Al_2O_3 = 102 g/mole

\text{ Mass of }Al_2O_3=(1.58moles)\times (102g/mole)=161.2g

Therefore, the mass of Al_2O_3 produced is, 161.2 grams.

Part (b) :

Now we have to calculate the percent yield of the reaction.

\text{Percent yield}=\frac{\text{Experimental yield}}{\text{Theoretical yield}}\times 100

Experimental yield = 116.9 g

Theoretical yield = 161.2 g

Now put all the given values in this formula, we get:

\text{Percent yield}=\frac{116.9g}{161.2g}\times 100=72.5\%

Therefore, the percent yield of the reaction is, 72.5 %

8 0
3 years ago
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