For the following incomplete combustion reaction, what are the coefficients necessary to balance the equation:
1 answer:
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The question is incomplete, here is the complete question:
How many grams of helium must be released to reduce the pressure to 65 atm assuming ideal gas behavior. Note : 334-mL cylinder for use in chemistry lectures contains 5.209 g of helium at 23°C.
<u>Answer:</u> The mass of helium released is 1.6 grams
<u>Explanation:</u>
We are given:
Mass of helium in the cylinder = 5.209 g
To calculate the number of moles, we use the equation given by ideal gas equation:
PV = nRT
Or,
where,
P = Pressure of the gas = 65 atm
V = Volume of the gas = 334 mL = 0.334 L (Conversion factor: 1 L = 1000 mL)
w = Weight of the gas = ?
M = Molar mass of helium gas = 4 g/mol
R = Gas constant =
T = Temperature of the gas =
Putting values in above equation, we get:
Mass of helium released = (5.209 - 3.573) g = 1.636 g = 1.6 g
Hence, the mass of helium released is 1.6 grams