Answer:
The correct answer is "32%".
Explanation:
The given values:
Weight of H,
= 4.9 g
Weight of sample,
= 15.8 g
Now,
The weight percentage of C will be:
=
By substituting the values, we get
=
=
In part 2 of the experiment, you will be analyzing a sample of household bleach. A 0.0800 g sample of household bleach is comple
tely reacted with KI(s). The resulting solution is then titrated with 0.150 M NaS2O3 solution. 0.500 mL of the solution is required to reach the colorless endpoint. What is the mass percent of NaOCl (MM = 74.44 g/mole) in the bleach?
1 answer:
Answer:
The mass percent of NaOCl = 3.48%
Explanation:
The mass percent of sodium hypochlorite in household bleach is determined by an oxidation-reduction titration with sodium thiosulfate.
In this reaction, the iodide is oxidized to form aqueous iodine, I₂ (aq) Iodine is not very soluble in water but in the presence of iodide ions, it immediately converts to the tri iodide ion (I₃⁻ ) which is soluble and forms a deep yellow solution.
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Answer:- Volume of the gas in the flask after the reaction is 156.0 L.
Solution:- The balanced equation for the combustion of ethane is:
From the balanced equation, ethane and oxygen react in 2:7 mol ratio or 2:7 volume ratio as we are assuming ideal behavior.
Let's see if any one of them is limiting by calculating the required volume of one for the other. Let's say we calculate required volume of oxygen for given 36.0 L of ethane as:
= 126 L
126 L of oxygen are required to react completely with 36.0 L of ethane but only 105.0 L of oxygen are available, It means oxygen is limiting reactant.
let's calculate the volumes of each product gas formed for 105.0 L of oxygen as:
= 60.0 L
Similarly, let's calculate the volume of water vapors formed:
= 90.0 L
Since ethane is present in excess, the remaining volume of it would also be present in the flask.
Let's first calculate how many liters of it were used to react with 105.0 L of oxygen and then subtract them from given volume of ethane to know it's remaining volume:
= 30.0 L
Excess volume of ethane = 36.0 L - 30.0 L = 6.0 L
Total volume of gas in the flask after reaction = 6.0 L + 60.0 L + 90.0 L = 156.0 L
Hence. the answer is 156.0 L.