A sample of hexane was cracked. A sample of 50cm³ of gas was produced. The gas decolorized 4cm³ of 0.23moldm–³ solution of bromine. What % of alkenes was produced?
millimoles Br2 used = mL x M = 4*0.23 = approx 0.9
1 mol Br2 adds to 1 mol alkene; therefore, mmols Br2 alkene = approx 0.9.
mols in 50 cc at stp = 50/22,400 and millimols = 50*1000/22,400 = approx 2.
% alkene = (mmols alk*100/mmols gas) = ?
% alk = mols alkene/
thanks Drbob222
To determine the percentage of alkenes produced in the sample of hexane, you need to calculate the amount of bromine consumed during the reaction.
Here's how you can do it step-by-step:
1. Determine the moles of bromine consumed:
- The volume of the bromine solution consumed is given as 4 cm³, and the concentration is 0.23 moldm⁻³.
- Convert the volume to liters: 4 cm³ = 4/1000 L = 0.004 L.
- Calculate the moles of bromine consumed using the formula: moles = concentration × volume.
- moles = 0.23 moldm⁻³ × 0.004 L = 0.00092 moles.
2. Calculate the moles of hexane:
- Given that the sample produced 50 cm³ of gas.
- Convert the volume to liters: 50 cm³ = 50/1000 L = 0.05 L.
- Since the gas produced is the product of cracking hexane, it contains the same number of moles as the hexane consumed.
- Therefore, the moles of hexane consumed are also 0.00092 moles.
3. Since hexane is an alkane and it produces only alkenes upon cracking, the moles of alkenes produced are equal to the moles of hexane consumed (0.00092 moles).
4. Calculate the percentage of alkenes produced:
- The percentage is calculated by dividing the moles of alkenes produced by the moles of hexane consumed and multiplying by 100.
- Percentage of alkenes = (moles of alkenes / moles of hexane) × 100.
- Percentage of alkenes = (0.00092 moles / 0.00092 moles) × 100 = 100%.
Therefore, 100% of alkenes were produced in the sample of hexane.