The volume of O₂ produced: 84.6 L
<h3>Further explanation</h3>
Given
7.93 mol of dinitrogen pentoxide
T = 48 + 273 = 321 K
P = 125 kPa = 1,23365 atm
Required
Volume of O₂
Solution
Decomposition reaction of dinitrogen pentoxide
2N₂O₅(g)→4NO₂(g)+O₂ (g)
From the equation, mol ratio N₂O₅ : O₂ = 2 : 1, so mol O₂ :
= 0.5 x mol N₂O₅
= 0.5 x 7.93
= 3.965 moles
The volume of O₂ :

First, calculating the amount of zinc carbonate by multiplying the given mass of the ore by the percent zinc carbonate.
mass of zinc carbonate = (976 tons)(0.78)
= 761.28 tons
Then, calculate for %Zn in zinc carbonate
%Zn = ((65.38) / (125.39)) x 100%
% Zn = 52.141%
tons of Zn from the sample,
mass of Zn = (761.28 tons)(0.52141)
= 396.94 tons Zn