Zinc and sulfur react to form zinc sulfide according to the equation Zn+S=>ZnS
a) How many grams of ZnS will be formed?
b) How many grams of the excess reactant will remain after the reaction is over?
Thank you!
To answer these questions, we first need to determine the limiting reactant and excess reactant. The limiting reactant is the reactant that is completely consumed in the reaction, while the excess reactant is the reactant that remains after the reaction is complete.
To find the limiting reactant, we compare the moles of each reactant to the balanced equation. The reactant that produces a smaller amount of product is the limiting reactant.
a) How many grams of ZnS will be formed?
Step 1: Convert the given mass of each reactant to moles.
To convert grams to moles, we need the molar mass of each element.
Molar mass of Zn = 65.38 g/mol
Molar mass of S = 32.06 g/mol
Assuming we have 10 grams of Zn and 20 grams of S:
Moles of Zn = mass of Zn / molar mass of Zn
= 10 g / 65.38 g/mol
≈ 0.1526 mol
Moles of S = mass of S / molar mass of S
= 20 g / 32.06 g/mol
≈ 0.6235 mol
Step 2: Determine the stoichiometric ratio between the reactants.
In the balanced equation Zn + S → ZnS, the ratio is 1:1. This means that 1 mole of Zn reacts with 1 mole of S to produce 1 mole of ZnS.
Step 3: Find the limiting reactant.
Since the ratio is 1:1, the limiting reactant is determined by comparing the moles of each reactant. In this case, we have fewer moles of Zn (0.1526 mol) compared to S (0.6235 mol), so Zn is the limiting reactant.
Step 4: Calculate the moles of ZnS produced.
Since the ratio is 1:1, the moles of ZnS produced will be equal to the moles of limiting reactant (Zn).
Moles of ZnS = moles of Zn = 0.1526 mol
Step 5: Convert moles of ZnS to grams.
To convert moles to grams, we multiply the moles by the molar mass of ZnS.
Molar mass of ZnS = molar mass of Zn + molar mass of S
= (65.38 g/mol) + (32.06 g/mol)
= 97.44 g/mol
Mass of ZnS = moles of ZnS × molar mass of ZnS
= 0.1526 mol × 97.44 g/mol
≈ 14.85 g
Therefore, approximately 14.85 grams of ZnS will be formed.
b) How many grams of the excess reactant will remain after the reaction is over?
Since Zn is the limiting reactant, it will be completely consumed. Therefore, none of the Zn will remain after the reaction.
To find the remaining amount of the excess reactant (S), we need to calculate the moles of S that reacted. This can be done by subtracting the moles of S used in the reaction from the initial moles of S.
Initial moles of S = moles of S in 20 grams of S (calculated in Step 1)
= 0.6235 mol
Moles of S used = moles of ZnS produced
= 0.1526 mol
Moles of S remaining = Initial moles of S - Moles of S used
= 0.6235 mol - 0.1526 mol
≈ 0.4709 mol
Finally, we can calculate the mass of S remaining by multiplying the moles of S remaining by the molar mass of S.
Mass of S remaining = moles of S remaining × molar mass of S
= 0.4709 mol × 32.06 g/mol
≈ 15.07 g
Therefore, approximately 15.07 grams of the excess reactant (S) will remain after the reaction is over.
Oh, sorry. 25 g of Zinc and 30 g of sulfur. I already found that Zinc is the limiting reagent, but I'm stuck on the rest.
If you know Zn is the limiting reagent then you must have moles Zn calculated as about 0.38.
Now you want to convert moles Zn to mols ZnS. To do that you use the coefficients in the balanced equation like this.
0.38 mol Zn x (1 mol ZnS/1 mol Zn) = 0.38 mol Zn x 1/1 so 0.38 mols Zn = 0.38 mol ZnS. You can convert that to grams by mols x molar mass = grams. (I just estimated mols Zn; I'm sure you have a better figure for that.)
To find S remaining S, do this. You have moles Zn. Convert to moles S (same process as converting mols Zn to mols ZnS. That will tell you how many moles S were used by the 25 g Zn. Then mols S x atomic mass S = grams S and that subtracted from 32 will tell you how many grams S did not react.