2ZnS + 3O2 -> 2ZnO + 2SO2

If 1000kg of ZnS is oxidised, what amount (in moles) and volume (in litres) of SO2 is produced (assuming that gas is at STP)

the formula for calculating the number of moles:

n=m/M
The formula for calculating the volume (at STP):
v=n.22,4
n(Zn)=1000.1000/97 moles
n(SO2)=n(Zn) = 1000.1000/97 moles
V(SO2) = 1000.1000.22,4/97=230927,8351 (in litres)

To determine the amount (in moles) and volume (in liters) of SO2 produced when 1000 kg of ZnS is oxidized, we need to follow these steps:

Step 1: Determine the molar mass of ZnS
The molar mass of ZnS (Zinc sulfide) can be calculated by adding the atomic masses of zinc (Zn) and sulfur (S).
Molar mass of ZnS = Atomic mass of Zn + Atomic mass of S
Molar mass of ZnS = (65.38 g/mol) + (32.06 g/mol)
Molar mass of ZnS = 97.44 g/mol

Step 2: Convert the mass of ZnS to moles
Given:
Mass of ZnS = 1000 kg = 1000,000 grams
Moles of ZnS = Mass of ZnS / Molar mass of ZnS
Moles of ZnS = 1000,000 g / 97.44 g/mol

Step 3: Use stoichiometry to determine the moles of SO2 produced
From the balanced chemical equation:
2 ZnS + 3 O2 → 2 ZnO + 2 SO2
The stoichiometric ratio between ZnS and SO2 is 2:2, meaning for every 2 moles of ZnS, 2 moles of SO2 are produced.
Therefore, moles of SO2 = Moles of ZnS

Step 4: Convert moles of SO2 to volume at STP
At standard temperature and pressure (STP), 1 mole of any gas occupies 22.4 liters.
So, the volume of SO2 at STP = Moles of SO2 x 22.4 L/mol

Substituting the values:
Moles of ZnS = 1000,000 g / 97.44 g/mol
Volume of SO2 at STP = Moles of ZnS x 22.4 L/mol

Calculating the values will give you the amount (in moles) and volume (in liters) of SO2 produced when 1000 kg of ZnS is oxidized.