Ammonia sulfate, an important fertilizer, can be prepared by the reaction of ammonia with sulfuric acid: 2NH3(g)+ H2SO4(aq) → (NH4)2SO4(aq) Calculate the volume of NH3(g) needed at 298 K and 25 atm to react with 1500 g of H2SO4.
Ammonia sulfate, an important fertilizer, can be prepared by the reaction of ammonia with sulfuric acid: 2NH3(g)+ H2SO4(aq) → (NH4)2SO4(aq) Calculate the volume of NH3(g) needed at 298 K and 25 atm to react with 1500 g of H2SO4.
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[tex]\large \mathbb {INSTRUCTION:}[/tex]
[tex]\sf 2NH_3(g+H_2SO_4(aq)→(NH_4)_2SO_4(aq)[/tex]
[tex] = = = = = = = = = = = = = = = = = = = = = = = = = = = = [/tex]
[tex]\large \mathbb {QUESTION:}[/tex]
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[tex]\large \mathbb {IDEAL \: GAS \: EQUATION:}[/tex]
[tex]\sf\:Volume=\frac{Moles×Gas \: constant×Temperature}{Pressure} [/tex]
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[tex]\large \mathbb {GIVEN \: DATA:}[/tex]
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[tex]\large \mathbb {SOLUTION:}[/tex]
The chemical reaction is shown below.
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The molar mass of sulfuric acid is 98.079 g/mol.
The moles of sulfuric acid can be calculated as shown below.
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Substitute the respective values in the above equation.
[tex] \sf \: Moles= \frac{Moles=150000g}{98.079g/mol} \\ \\ \sf \boxed{ \sf \: =1529.38mol}[/tex]
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According to the above reaction, 1 mole of sulfuric acid reacts with 2 moles of ammonia.
So, the moles of ammonia are [tex]\sf \: 2 \times 1529.38mol=3058.76mol.[/tex]
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The volume of ammonia can be calculated as shown below.
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The value of gas constant is 0.0821 L.atm/mol.K.
Substitute the respective values in the above equation.
[tex]\sf \boxed {= \:2943L}[/tex]
Hence the answer is: 2943 L
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What is the Ideal Gas Law:
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Ammonia sulfate, an important fertilizer, can be prepared by the reaction of ammonia with sulfuric acid: 2NH3(g)+ H2SO4(aq) → (NH4)2SO4(aq) Calculate the volume of NH3(g) needed at 298 K and 25 atm to react with 1500 g of H2SO4.
the answer is 2943 L
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