## anonymous one year ago PLS HELP!!!! which volume of oxygen measured at room temperature and pressure is needed for complete combustion of 0.1 mol of propan-1-ol? a)10.8dm3 b)12.0dm3 c)21.6dm3 d)24.0dm3 answer is A...

1. Ciarán95

Given any organic compound, it will combust (i.e. react with oxygen) to produce carbon dioxide (CO2) and water (H2O). The balanced chemical equation for the combustion of propan-1-ol (CH3CH2CH2OH, or C3H7OH) in air is as follows: $2C _{3}H _{7}OH + 9O _{2} \rightarrow 6CO _{2} + 8H _{2}O$ From looking at the coefficients, this means that for every 2 moles of propan-1-ol we have, it will react with 9 moles of oxygen gas to produce 6 moles of carbon dioxide and 8 moles of water. We also assume that propan-1-ol is the limiting reagent to this reaction and that the number of moles of the organic compound we have dictates the amount of products formed (for combusion reactions, we generally always assume that there is excess oxygen present for it to take place). In this question however, were looking to find the minimum volume of oxygen that must be present to react with 0.1 moles of propan-1-ol to form the products, given the coefficients appearing in the balanced chemical equation. So, propan-1-ol will react with O2 gas in a 2:9 molar ratio, or a 1:9/2 ratio when rewritten. Given 0.1 moles of the organic compound therefore, we would require (0.1)(9/2) moles of O2 to be present. From the Ideal Gas equation, 1 mole of any ideal substance in the gaseous state will occupy 22.4 litres at standard temperature and pressure, or 22.4 dm3. So, having worked out the number of moles O2 gas required above, we simply multiply this value by 22.4 to enable us to convert to the final volume of oxygen required for complete combustion.

2. Ciarán95

The answer I actually got was 10.08 dm3 when working this out...it's closest to A but not exactly matching it. What value did you get out when you calculated it?

3. JoannaBlackwelder

I agree with @Ciarán95 for everything except I get the molar volume of gas at room temp to be 24 L/mol (the value you gave was for STP).

4. JoannaBlackwelder

@Naheeda

5. anonymous

thank you!! @JoannaBlackwelder and @Ciarán95... I got the same... I guess my equation was wrong... thanx.

6. Ciarán95

No problem @Naheeda - apologies for the mix-up over the molar volume of gas at room tempearature as opposed to STP. @JoannaBlackwelder you were absolutely right about that one!