Question

Question #6da34

Answer 1

`=0.12molAlCl_3`

Explanation:

1. Since data are already in moles, find the the required moles of each substanced considering a complete reaction to find the limiting reactant; i.e.,
   `A. etaAl=0.15mol`
   `=0.15cancel(molAl)xx(6molHCl)/(2cancel(molAl))`
   `=0.45molHCl`

`"This means that:"`
`0.15molAl-=0.45molHCl`

`"Therefore"`

`(etaHCl " available")/(=0.35molHCl)<(etaHCl " required")/(=0.45molHCl)`

`B. etaHCl=0.35mol`
`=0.35cancel(molHCl)xx(2molAl)/(6cancel(molHCl))`
`=0.12molAl`

`"This means that:"`
`0.35molHCl-=0.1167molAl`

`"Therefore"`

`(etaAl " available")/(=0.15molAl)>(etaAl " required")/(=0.12molHCl)`

1. Thus, the limiting reactant is `HCl`. Now. find the `AlCl_3` formed; i.e.,
   `=0.35cancel(molHCl)xx(2molAlCl_3)/(6cancel(molHCl))`
   `=0.12molAlCl_3`