If 2.0 Grams of anhydrous sodium sulfate are mixed with 2.0 grams of calcium chloride dihydrate and 2.0 grams of calcium sulfate dihydrate are collected, what is the percent yield? DO NOT ROUND A NUMBER IF USED IN A SUBSEQUENT CALCULATION
First you need to find the limited reactant. You do this by dividing the 2.0 g of NaSO4 by it's molar mass as well as the 2.0 g of Calcium Sulfate dihyrdate (CaCl2 + 2H20) You'll find that the Calcium sulfate is the limited reactant. From there you Take the moles of Calcium chloride dihydrate(which you just found because you found it to be the limited reactant) and times it by the molar mass of Calcium Sulfate dihydrate. From there you get the theoretical yield. Now you just divide the actual by the theoretical. so 2.0/2.342 (grams) to get approximately 85%
Hey tanner, you did your math wrong and you confused a bunch of words in your paragraph. Double check a post before you submit it buddy.
I actually found tanners math to be great. Some of the words were off though.
To find the percent yield, we need to compare the actual yield (the amount of calcium sulfate dihydrate collected) to the theoretical yield (the maximum amount of calcium sulfate dihydrate that could be produced based on the balanced chemical equation).
First, let's determine the balanced chemical equation for the reaction between anhydrous sodium sulfate and calcium chloride dihydrate to form calcium sulfate dihydrate:
Na2SO4 + CaCl2 • 2H2O --> CaSO4 • 2H2O + 2NaCl
According to the balanced equation, 1 mole of anhydrous sodium sulfate reacts with 1 mole of calcium chloride dihydrate to produce 1 mole of calcium sulfate dihydrate.
We need to calculate the number of moles of anhydrous sodium sulfate and calcium chloride dihydrate based on their given masses:
For anhydrous sodium sulfate:
- Given mass: 2.0 grams
- Molar mass of Na2SO4: approximately 142.04 g/mol
- Number of moles = mass / molar mass = 2.0 g / 142.04 g/mol = 0.0141 mol
For calcium chloride dihydrate:
- Given mass: 2.0 grams
- Molar mass of CaCl2 • 2H2O: approximately 147.01 g/mol
- Number of moles = mass / molar mass = 2.0 g / 147.01 g/mol = 0.0136 mol
Since the balanced chemical equation shows a 1:1 mole ratio between anhydrous sodium sulfate and calcium sulfate dihydrate, the smaller number of moles (0.0136 mol) will be the limiting reactant. This means it will determine the maximum amount of calcium sulfate dihydrate that can be produced.
The theoretical yield of calcium sulfate dihydrate can be calculated based on the number of moles of the limiting reactant:
- Theoretical yield of CaSO4 • 2H2O = moles of limiting reactant × molar mass of CaSO4 • 2H2O
= 0.0136 mol × 172.18 g/mol (molar mass of CaSO4 • 2H2O) = 2.34 grams
Now, we can calculate the percent yield by dividing the actual yield by the theoretical yield and multiplying by 100:
- Percent yield = (actual yield / theoretical yield) × 100
= (2.0 g / 2.34 g) × 100
= 85.47%
Therefore, the percent yield of calcium sulfate dihydrate in this reaction is 85.47%.