Zinc sulfide and oxygen gas react to form oxide and sulfur dioxide. Determine the amount of ZnO that should be produced in a reaction between 46.5 g of ZnS and 13.3 g of oxygen. What is the mass of the xs reactant? Please show work
Do you know how to work a simple stoichiometry problem; i.e., say 46.5 g ZnS and ALL OF THE oxygen you need (in other words don't complicate it with an amount of oxygen)?
Here is one worked. If you know how to do that, the limiting reagent problems are easy.
http://www.jiskha.com/science/chemistry/stoichiometry.html
To determine the amount of ZnO produced in the reaction, we need to find the limiting reactant. The limiting reactant is the reactant that is completely consumed and determines the maximum amount of product that can be formed.
First, we need to write a balanced chemical equation for the reaction:
2 ZnS + 3 O2 → 2 ZnO + 2 SO2
According to the balanced equation, the stoichiometric ratio between ZnS and ZnO is 2:2 (or 1:1), meaning that 1 mole of ZnS reacts with 1 mole of ZnO.
1 mole of ZnS has a molar mass of 97.45 g (the atomic masses of Zn and S are 65.38 g/mol and 32.06 g/mol respectively).
Using the given mass of ZnS (46.5 g), we can calculate the number of moles of ZnS:
moles of ZnS = mass of ZnS / molar mass of ZnS
moles of ZnS = 46.5 g / 97.45 g/mol
Next, we need to determine the amount of oxygen gas (O2) in terms of moles. Oxygen gas has a molar mass of 32.00 g/mol.
Using the given mass of oxygen (13.3 g), we can calculate the number of moles of oxygen:
moles of O2 = mass of O2 / molar mass of O2
moles of O2 = 13.3 g / 32.00 g/mol
Now, we compare the moles of ZnS and O2 to determine the limiting reactant.
The ratio between ZnS and O2 is 2:3 (from the balanced equation).
If the moles of ZnS and O2 are already in the 2:3 ratio, then whichever reactant has fewer moles is the limiting reactant. In other words, the reactant with a smaller moles-to-ratio ratio will determine the amount of product formed.
By comparing the moles of ZnS and O2, in this case, we can see that the moles of ZnS is 0.477 moles and the moles of O2 is 0.415625 moles.
Since the moles of ZnS (0.477) are greater than the moles of O2 (0.415625), O2 is the limiting reactant.
To find the amount of ZnO produced, we use the stoichiometry from the balanced equation. The stoichiometric ratio between ZnO and O2 is 2:3, meaning that 2 moles of ZnO are formed for every 3 moles of O2.
moles of ZnO = (moles of O2) x (2 moles of ZnO / 3 moles of O2)
moles of ZnO = 0.415625 moles x (2/3)
Finally, to find the mass of ZnO, we use the molar mass of ZnO, which is 81.38 g/mol:
mass of ZnO = (moles of ZnO) x (molar mass of ZnO)
mass of ZnO = (0.415625 moles) x (81.38 g/mol)
By calculating the above expression, we find the mass of ZnO to be approximately 27.826 g.
To determine the mass of the excess reactant (the reactant that is not completely consumed), we subtract the mass of the limiting reactant from the given mass of the excess reactant.
Mass of excess reactant (O2) = given mass of O2 - mass of O2 used in the reaction
Mass of excess reactant (O2) = 13.3 g - (moles of O2 used x molar mass of O2)
We already calculated that 0.415625 moles of O2 were used in the reaction, so we can substitute that value:
Mass of excess reactant (O2) = 13.3 g - (0.415625 moles x 32.00 g/mol)
By calculating the above expression, we find the mass of the excess reactant (O2) to be approximately 0.769 g.