CHEMISTRY HIGH SCHOOL

Ammonia and sulfuric acid react to form ammonium sulfate. Determine the starting mass of each reactant if 20.3 g of ammonium sulfate is produced and 5.89g of sulfuric acid remains unreacted.

Answers

Answer 1

Answer:

Answer: The mass of ammonia is 5.236 g and that of sulfuric acid is 15.064 g

Explanation:

Calculating the mass of ammonia:

To calculate the number of moles, we use the equation:

$\text{Number of moles}=\frac{\text{Given mass}}{\text{Molar mass}}$ .....(1)

Given mass of ammonium sulfate = 20.3 g

Molar mass of ammonium sulfate = 132.14 g/mol

Putting values in equation 1, we get:

$\text{Moles of ammonium sulfate}=(20.3g)/(132.14g/mol)=0.154mol$

The chemical equation for the reaction of ammonia and sulfuric acid follows:

$2NH_3+H_2SO_4\rightarrow (NH_4)_2SO_4$

As, sulfuric acid remains unreacted, which means it is an excess reagent and its starting mass cannot be determined from ammonium sulfate.

By Stoichiometry of the reaction:

1 mole of ammonium sulfate is produced by 2 moles of ammonia.

So, 0.154 moles of ammonium sulfate is produced by = $(2)/(1)* 0.154=0.308mol$ of ammonia.

Now, calculating the mass of ammonia from equation 1, we get:

Molar mass of ammonia = 17 g/mol

Moles of ammonia = 0.308 moles

Putting values in equation 1, we get:

$0.308mol=\frac{\text{Mass of ammonia}}{17g/mol}\n\n\text{Mass of ammonia}=5.236g$

Calculating the mass of sulfuric acid

Law of conservation of mass states that mass can neither be created nor be destroyed but it can only be transformed from one form to another form.

This also means that total mass on the reactant side must be equal to the total mass on the product side.

Let the mass of sulfuric acid be 'x' grams

We are given:

Mass of ammonium sulfate = 20.3 grams

Mass of ammonia = 5.236 grams

Total mass on reactant side = 5.236 + x

Total mass on product side = 20.3 g

So, by applying law of conservation of mass, we get:

$5.236+x=20.3\n\nx=15.064g$

Hence, the mass of ammonia is 5.236 g and that of sulfuric acid is 15.064 g

Answer 2

Answer: 2NH₃ + H₂SO₄ ⇒ (NH₄)₂SO₄
34.06g : 98.08g : 132.14g
x y 20.03g

x = [20.03g*34.06g]/132.14g = 5.16g

5.16g + y = 20.03g ⇔ from stoichiometric ratio
20.03g - 5.16g = y
y = 14,87g + 5.89g = 20.76g ⇔ 5.89g of sulfuric(VI) acid remains unreacted

Starting mass:

mNH₃ = 5.16g
mH₂SO₄ = 20.76g

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Answers

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Explanation :

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