For the reaction
? ZnS + ? O2 →? ZnO+? SO2 , what is the maximum amount of ZnO which could be formed from 14.06 g of ZnS and 5.56 mol of O2?
Answer in units of g
A limiting reagent problem.
To find the maximum amount of ZnO that can be formed, we need to determine the limiting reactant. The limiting reactant is the reactant that is completely consumed and limits the amount of product formed.
First, we need to determine the number of moles of ZnS and O2 in the given quantities.
The molar mass of ZnS (Zinc Sulfide) is 97.44 g/mol. Therefore,
Number of moles of ZnS = mass of ZnS / molar mass of ZnS
= 14.06 g / 97.44 g/mol
Next, we need to determine the number of moles of O2. We know that there are 5.56 mol of O2.
Now, we will use the balanced chemical equation to determine the stoichiometry of the reaction:
? ZnS + ? O2 → ? ZnO + ? SO2
From the balanced equation, we can see that the stoichiometric ratio between ZnS and ZnO is 1:1. This means that for every 1 mole of ZnS reacted, we get 1 mole of ZnO.
To calculate the maximum amount of ZnO formed, we need to determine the limiting reactant. This can be done by comparing the mole ratios of ZnS and O2 in the reaction with the actual moles of ZnS and O2 present.
The mole ratio of ZnS to ZnO is 1:1, and the moles of ZnS is calculated above.
Therefore, the maximum moles of ZnO formed = moles of ZnS
To calculate the maximum amount of ZnO formed in grams, we multiply the maximum moles of ZnO by its molar mass.
Molar mass of ZnO (Zinc Oxide) = 81.37 g/mol
Maximum amount of ZnO formed = maximum moles of ZnO * molar mass of ZnO
= moles of ZnS * molar mass of ZnO
Finally, substitute the moles of ZnS calculated earlier:
Maximum amount of ZnO formed = (14.06 g / 97.44 g/mol) * 81.37 g/mol
Evaluate the above expression to find the maximum amount of ZnO formed, and express the answer in units of grams.