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Veronika [31]
3 years ago
14

Ammonium sulfate (NH4)2SO4 is made by reacting 25.0 L of 3.0 mol/L H2SO4 with 3.1× 103 L of NH3 at a pressure of 0.68 atm and a

temperature of 298 K according to the following reaction .
NH3(g) + H2SO4(aq) → (NH4)2SO4 (aq)

How many grams of ammonium sulfate are produced?
Chemistry
1 answer:
White raven [17]3 years ago
5 0

<u>Answer:</u> The mass of (NH_4)_2SO_4 produced is 9910.5 g

<u>Explanation:</u>

  • <u>For </u>H_2SO_4<u>:</u>

Molarity is calculated by using the equation:

\text{Molarity}=\frac{\text{Moles}}{\text{Volume}} ......(1)  

Molarity of H_2SO_4 = 3.0 M

Volume of solution = 25.0 L

Putting values in equation 1, we get:

\text{Moles of }H_2SO_4=(3.0mol/L\times 25.0L)=75mol

  • <u>For </u>NH_3<u>:</u>

The ideal gas equation is given as:

PV=nRT .......(2)

where,

P = pressure of the gas = 0.68 atm

V = volume of gas = 3.1\times 10^3L

n = number of moles of gas = ? moles

R = Gas constant = 0.0821 L.atm/mol.K

T = temperature of the gas = 298 K

Putting values in equation 2, we get:

0.68atm\times 3.1\times 10^3L=n\times 0.0821L.atm/mol.K\times 298K\\\\n=\frac{0.68\times 3.1\times 10^3}{0.0821\times 298}=86.16mol

For the given chemical equation:

NH_3(g)+H_2SO_4(aq)\rightarrow (NH_4)_2SO_4(aq)

By stoichiometry of the reaction:

If 1 mole of H_2SO_4 reacts with 1 mole of NH_3

So, 75 moles of H_2SO_4 will react with = \frac{1}{1}\times 75=75mol of NH_3

As the given amount of NH_3 is more than the required amount. Thus, it is present in excess and is considered as an excess reagent

Thus, H_2SO_4 is considered a limiting reagent because it limits the formation of the product.

By the stoichiometry of the reaction:

If 1 mole of H_2SO_4 produces 1 mole of (NH_4)_2SO_4

So, 75 moles of H_2SO_4 will produce = \frac{1}{1}\times 75=75mol of (NH_4)_2SO_4

The number of moles is defined as the ratio of the mass of a substance to its molar mass. The equation used is:

\text{Number of moles}=\frac{\text{Given mass}}{\text{Molar mass}}

We know, molar mass of (NH_4)_2SO_4 = 132.14 g/mol

Putting values in above equation, we get:

\text{Mass of }(NH_4)_2SO_4=(75mol\times 132.14g/mol)=9910.5g

Hence, the mass of (NH_4)_2SO_4 produced is 9910.5 g

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