In previous chapters when doing percent yield problems, Iconverted moles to grams before plugging it into the percent yieldformula. Why is it done differently on this problem? If 5.97g of C6H12O6 isreacted and 1.44L of CO2 gas are collected at 293K and0.984 atm, what is the percent yield of the reaction? C6H12O6 (s) yields2C2H5OH (l) + 2CO2 (g) answer was 88.8% In previous chapters when doing percent yield problems, Iconverted moles to grams before plugging it into the percent yieldformula. Why is it done differently on this problem? If 5.97g of C6H12O6 isreacted and 1.44L of CO2 gas are collected at 293K and0.984 atm, what is the percent yield of the reaction? C6H12O6 (s) yields2C2H5OH (l) + 2CO2 (g) answer was 88.8%
Explanation / Answer
C6H12O6(s) ..................> 2C2H5OH(l) + 2CO2 (g) Moles = mass / molar mass So moles of C6H12O6(s) is =5.97g / ( 180.16g/mol) =0.0331mol Moles of CO2 = PV /RT = (0.984atm)(1.44L) / (0.0821Latm mol^-1 K^-1)(293K) = 0.0589 mol Moles of C6H12O6 (s)get reacted = (1mol C6H12O6 (s) /2mol CO2 (g) ) * 0.0589mol CO2(g) = 0.0295mol 0.0295mol of C6H12O6 (s) = 0.0295mol *(180. 16g /mol) = 5.31g Percent yeild = ( 5.31g / 5.97g) * 100 = 88.94% C6H12O6(s) ..................> 2C2H5OH(l) + 2CO2 (g) Moles = mass / molar mass So moles of C6H12O6(s) is =5.97g / ( 180.16g/mol) =0.0331mol Moles of CO2 = PV /RT = (0.984atm)(1.44L) / (0.0821Latm mol^-1 K^-1)(293K) = 0.0589 mol Moles of C6H12O6 (s)get reacted = (1mol C6H12O6 (s) /2mol CO2 (g) ) * 0.0589mol CO2(g) = 0.0295mol 0.0295mol of C6H12O6 (s) = 0.0295mol *(180. 16g /mol) = 5.31g Percent yeild = ( 5.31g / 5.97g) * 100 = 88.94%