Question

Question #7607c

Answer 1

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%`