Question #7607c

Question

912 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2015-03-17T00:41:47

We do all this by ratios #2H_2+O_2->2H_2O#

Molecular mass:
#H_2->2*1=2# times 2= #4#
#O_2->2*16=32#
#H_2O->2*1+1*16=18# times 2= #36#
So the weight ratio of this reaction is
#4+32->36or1+8->9#

Now we calculate how much #O_2# is needed for the #H_2#:
#16.8/1=x/8->x=134.4g# which we don't have.

The other way around:
#x/1=46.4/8->x=5.8g#

So #5.8g# of #H_2# will react with all #46.4g# of #O_2#

This should give #5.8+46.4=52.2g# of #H_2O#

The percentage yield is thus:
#12/52.2 *100~~23%#