Question

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?

Answer 1

Answer: The mass of $(NH_4)_2SO_4$ produced is 9910.5 g

Explanation:

• For $H_2SO_4$:

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:

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

• For $NH_3$:

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