# Question #fd7ab

Mar 17, 2017

First convert weight to moles

$\text{Moles" = "weight"/"molar mass}$

Therefore moles in 25g of aliminium

= $\text{25g"/"26.981539g/mol" = "0.926moles}$

If you had excess of chlorine

The amount of $A l C {l}_{3}$ formed would be 0.926moles because the molar ratio of aluminium to aluminium trichloride = 2 : 2 = 1 : 1

If there would be 30g of chlorine then the amount of $A l C {l}_{3}$ formed =

Moles in 30g of chlorine

= $\text{30g"/(35.453 * 2)= "0.42309536569moles}$

Balanced equation

= $2 A l + 3 C {l}_{2} = 2 A l C {l}_{3}$

According to this aluminium is excess

Therefore aluminium needed$= \text{ 0.4231mol"/("3mol" xx "2mol") = 0.28206357713"moles}$

Ratio of aluminium to aluminium trichloride

= 1 : 1

Therefore aluminium trichloride formed is 0.28206357713moles