a. 43.1 g
b. 38.2%
<h3>Further explanation</h3>
Given
32.5 grams of NaOH
Required
The theoretical yield of Na₂CO₃
The percent yield
Solution
Reaction
2NaOH(s) + CO₂(g) → Na₂CO₃(s) + H₂O(l)
mol NaOH :
= mass : MW
= 32.5 : 40 g/mol
= 0.8125
mol Na₂CO₃ from the equation :
= 1/2 x mol NaOH
= 1/2 x 0.8125
= 0.40625
a.
Mass Na₂CO₃ :
= mol x MW Na₂CO₃
= 0.40625 x 106 g/mol
= 43.0625≈43.1 g
b. % yield = (actual/theoretical) x 100%
%yield = 16.45/43.1 x 1005
%yield = 38.17%≈38.2%
24 gFeF3 x (1 mol FeF3/grams FeF3)
x (6.02x10^23 molecules FeF3/ 1 mol FeF3)
Just Calculate Molar Mass of FeF3 and plug into equation
The part of the experiment that’s is not touched by the independent variable and is for comparison is called the :
Control Group
Answer:
The mass percent of aluminum sulfate in the sample is 16.18%.
Explanation:
Mass of the sample = 1.45 g
Mass of the precipitate = 0.107 g
Moles of aluminum hydroxide = 
According to reaction, 2 moles of aluminum hydroxide is obtained from 1 mole of aluminum sulfate .
Then 0.001372 moles of aluminum hydroxide will be obtained from:
Mass of 0.000686 moles of aluminum sulfate :
= 0.000686 mol × 342 g/mol = 0.2346 g
The mass percent of aluminum sulfate in the sample:
Answer:
Random samples
Explanation:
It needs to be random so that there isn't bias that would skew the consistency
