A)Three gases (8.00 g of methane,CH 4 , 18.0 g of ethane, C 2 H 6 , andan unknow
ID: 686980 • Letter: A
Question
A)Three gases (8.00 g of methane,CH4, 18.0 g of ethane, C2H6, andan unknown amount of propane, C3H8) wereadded to the same 10.0-L container. At 23.0 oC, thetotal pressure in the container is 3.70 atm. Calculate thepartial pressure of each gas in the container. Express the pressure valuesnumerically in atmospheres, separated by commas. Enter the partialpressure of methane first, then ethane, then propane. B) A gaseous mixture of O2 andN2 contains 40.8 % nitrogen by mass. What is thepartial pressure of oxygen in the mixture if the total pressure is525 mmHg? A)Three gases (8.00 g of methane,CH4, 18.0 g of ethane, C2H6, andan unknown amount of propane, C3H8) wereadded to the same 10.0-L container. At 23.0 oC, thetotal pressure in the container is 3.70 atm. Calculate thepartial pressure of each gas in the container. Express the pressure valuesnumerically in atmospheres, separated by commas. Enter the partialpressure of methane first, then ethane, then propane. B) A gaseous mixture of O2 andN2 contains 40.8 % nitrogen by mass. What is thepartial pressure of oxygen in the mixture if the total pressure is525 mmHg? A gaseous mixture of O2 andN2 contains 40.8 % nitrogen by mass. What is thepartial pressure of oxygen in the mixture if the total pressure is525 mmHg?Explanation / Answer
We Know that : According to ideal gas equation: PV = nRT 3.70 atm x10.0 L = n x 0.0821 atm-L/mol-K x 296 K numberof moles of gases = 1.522 mol The number of molesof CH4 = 8 / 16 = 0.5 mol number of moles ofC2H6 = 18 / 30 = 0.6 mol number ofmoles of propane = 1.522 - ( 0.5 +0.6 ) = 0.422 mol weightof the propane gas taken = 0.422 mol x 44 g / mol = 18.568 g Partial pressure of theCH4 gas = 3.70 atm x 0.328 = 1.2155 atm Partial pressureof C2H6 gas = 3.70 atm x0.3942 = 1.458 atm Partial pressureof C3H8 gas = 3.70 atm x0.277 = 1.025 atmRelated Questions
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