Can someone please show me how to do this question,
Thank you


2005 - Written Paper - Q3(b)(ii)
Started by AmAnDa, May 23 2006 02:45 PM
3 replies to this topic
#1
Posted 23 May 2006 - 02:45 PM
(`'•.¸(`'•.¸ ¸.•'´)¸.•'´)
x..*...x.°•.♥.•°.Amanda loves Craig.°•.♥.•°.x...*..x
(¸.•'´(¸.•'´ `'•.¸)`'•.¸)
#2
Posted 23 May 2006 - 03:13 PM
Ok here it goes:
1 mole ethanol -----> 1 mole ethyl etanoate
46g -------->88g
1g--------> 88/46 g
5g ------> 88/46 x 5 = 9.565g
% yield = actual yield / theoretical yield x 100
% yield = 5.8/9.565 x 100
% yield = 60.6% (note the answers round it up to 61%)
1 mole ethanol -----> 1 mole ethyl etanoate
46g -------->88g
1g--------> 88/46 g
5g ------> 88/46 x 5 = 9.565g
% yield = actual yield / theoretical yield x 100
% yield = 5.8/9.565 x 100
% yield = 60.6% (note the answers round it up to 61%)
#3
Posted 23 May 2006 - 03:16 PM
why does everyone have such a problem with yield??
i thought it was one of the easier calculations in higher chemistry
i thought it was one of the easier calculations in higher chemistry

#4
Posted 24 May 2006 - 11:07 AM
Ok here it goes:
1 mole ethanol -----> 1 mole ethyl etanoate
46g -------->88g
1g--------> 88/46 g
5g ------> 88/46 x 5 = 9.565g
% yield = actual yield / theoretical yield x 100
% yield = 5.8/9.565 x 100
% yield = 60.6% (note the answers round it up to 61%)
Thank you, much appreciated.

(`'•.¸(`'•.¸ ¸.•'´)¸.•'´)
x..*...x.°•.♥.•°.Amanda loves Craig.°•.♥.•°.x...*..x
(¸.•'´(¸.•'´ `'•.¸)`'•.¸)
1 user(s) are reading this topic
0 members, 1 guests, 0 anonymous users