Answer: The mass of [tex](NH_4)_2SO_4[/tex] produced is 9910.5 g
Explanation:
Molarity is calculated by using the equation:
[tex]\text{Molarity}=\frac{\text{Moles}}{\text{Volume}}[/tex] ......(1)
Molarity of [tex]H_2SO_4[/tex] = 3.0 M
Volume of solution = 25.0 L
Putting values in equation 1, we get:
[tex]\text{Moles of }H_2SO_4=(3.0mol/L\times 25.0L)=75mol[/tex]
The ideal gas equation is given as:
[tex]PV=nRT[/tex] .......(2)
where,
P = pressure of the gas = 0.68 atm
V = volume of gas = [tex]3.1\times 10^3L[/tex]
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:
[tex]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[/tex]
For the given chemical equation:
[tex]NH_3(g)+H_2SO_4(aq)\rightarrow (NH_4)_2SO_4(aq)[/tex]
By stoichiometry of the reaction:
If 1 mole of [tex]H_2SO_4[/tex] reacts with 1 mole of [tex]NH_3[/tex]
So, 75 moles of [tex]H_2SO_4[/tex] will react with = [tex]\frac{1}{1}\times 75=75mol[/tex] of [tex]NH_3[/tex]
As the given amount of [tex]NH_3[/tex] is more than the required amount. Thus, it is present in excess and is considered as an excess reagent
Thus, [tex]H_2SO_4[/tex] is considered a limiting reagent because it limits the formation of the product.
By the stoichiometry of the reaction:
If 1 mole of [tex]H_2SO_4[/tex] produces 1 mole of [tex](NH_4)_2SO_4[/tex]
So, 75 moles of [tex]H_2SO_4[/tex] will produce = [tex]\frac{1}{1}\times 75=75mol[/tex] of [tex](NH_4)_2SO_4[/tex]
The number of moles is defined as the ratio of the mass of a substance to its molar mass. The equation used is:
[tex]\text{Number of moles}=\frac{\text{Given mass}}{\text{Molar mass}}[/tex]
We know, molar mass of [tex](NH_4)_2SO_4[/tex] = 132.14 g/mol
Putting values in above equation, we get:
[tex]\text{Mass of }(NH_4)_2SO_4=(75mol\times 132.14g/mol)=9910.5g[/tex]
Hence, the mass of [tex](NH_4)_2SO_4[/tex] produced is 9910.5 g