Example 4, Tulsa Community × M A sample of an ideal 1 m tulsaccedu | _ Blackboar

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Example 4, Tulsa Community × M A sample of an ideal 1 m tulsaccedu | _ Blackboard Learn php?id-3884919 rning.co Sapling Learning The combustion of octane, Cet. proceeds according to the reaction 25olg) 16C0,(g)+18H20() If 466 mol of octane combusts, what volume of carbon dioxide is produced at 35.0 and 0.995 atm? Number 99501.025 L O Previous Give Up & View Solution Try Again Next Explanation First, find out the number of moles of CO2 produced from the combustion of 466 mol CeHs. The coefficients of the chemical equation specify that for every 2 mol of CsH. 16 mol of CO2 is produced. Then use the ideal gas equation, PV= nRT to solve for V. Make sure that you have converted the temperature into kelvin before using the ideal gas equation. ch Up

Explanation / Answer

From the balanced equation we can say that

2 mole of octane produces 16 mole of CO2 so

466 mole of octane will produce

= 466 mole of octane *(16 mole of CO2 / 2 mole of octane)

= 3728 mole of CO2

PV = nRT

V = nRT/P = (3728*0.0821*308) / 0.995 = 94269 / 0.995 = 94742.905 L

Therefore, the volume is 94742.905 L

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