A 0.271 g sample of an unknown vapor occupies 294 ml at 140.°c and 847 mmhg. the empirical formula of the compound is ch2. what
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<u>Answer:</u> The volume of chlorine gas produced in the reaction is 2.06 L.
<u>Explanation:</u>
- <u>For potassium permanganate:</u>
To calculate the number of moles, we use the equation:
Given mass of potassium permanganate = 6.23 g
Molar mass of potassium permanganate = 158.034 g/mol
Putting values in above equation, we get:
- <u>For hydrochloric acid:</u>
To calculate the moles of hydrochloric acid, we use the equation:
Molarity of HCl = 6.00 M
Volume of HCl = 45.0 mL = 0.045 L (Conversion factor: 1 L = 1000 mL)
Putting values in above equation, we get:
- For the reaction of potassium permanganate and hydrochloric acid, the equation follows:
By Stoichiometry of the reaction:
16 moles of hydrochloric acid reacts with 2 moles of potassium permanganate.
So, 0.27 moles of hydrochloric acid will react with = of potassium permanganate.
As, given amount of potassium permanganate is more than the required amount. So, it is considered as an excess reagent.
Thus, hydrochloric acid is considered as a limiting reagent because it limits the formation of product.
By Stoichiometry of the reaction:
16 moles of hydrochloric acid reacts with 5 moles of chlorine gas.
So, 0.27 moles of hydrochloric acid will react with = of chlorine gas.
- To calculate the volume of gas, we use the equation given by ideal gas equation:
where,
P = pressure of the gas = 1.05 atm
V = Volume of gas = ? L
n = Number of moles = 0.0843 mol
R = Gas constant =
T = temperature of the gas =
Putting values in above equation, we get:
Hence, the volume of chlorine gas produced in the reaction is 2.06 L.