Answer:
a. how much So² can be made in moles?
= 0.2 moles
b. how much So² can be made in grams?
= 12.8 grams
c. what is the Excess reactant?
= O²
Step-by-step explanation:
mass of ZnS= 19.6g
mass of O²= 33.6g
i. start by determining if the equation is balanced (it is)
ii. calculate the molar mass of each reagent
- molar mass of ZnS
= (65) + (32) = 97g/mol
- molar mass of O²
= (16 x 2) = 32g/mol
iii. determine the reactants' number of moles
- no of moles of ZnS = mass of ZnS / molar mass of ZnS
= 19.6 / 97
= 0.2 mol
- no of moles of O² = mass of O² / molar mass of O²
= 33.6 / 32
= 1.05 mol
iv. determine the reactant mole ratio
0.2 moles of ZnS reacted with 1.05 moles of O²
ratio = 1.05 / 0.2 which will approximately be 5
therefore, we can say that there are x5 as many molecules of O² than ZnS
v. determine the reactant ideal ratio
from the equation, 2 moles of ZnS reacted with 3 moles of 0². therefore, the ideal ratio is 2:3, with O² having the greater ratio once again
vi. determine the limiting and excess reactant
we can say from both the mole and ideal ratio that O² is the excess reactant and ZnS is the limiting reactant (answer to the third question)
next phase,
vii. determine the ratio of limiting reactant to the product
from the equation,
2 moles of ZnS produced 2 moles of So²
therefore, mole ratio = 2:2 which is also 1:1
viii. to determine how much So² can be made in moles
= no of moles of limiting reactant x mole ratio of reactant and product
= 0.2 moles of ZnS x 1/1
= 0.2 x 1
= 0.2 moles of So² (answer to the first question)
ix. to determine how much So² can be made in grams
= no of moles of So² produced x molar mass of So²
- molar mass of So²
= 32 + (16 x 2)
= 32 + 32
= 64g/mol
therefore, how much So² produced in grams
= 0.2 x 64
= 12.8g of So² (answer to the second question)
answerer's note: I recently started learning about theoretical yield in school and so this is the result of what I know so far. You're free to make confirmations from others who are more experienced :)