If you start with 4.5 moles of aluminum and 6.5 moles of copper chloride to make aluminum chloride and copper, what is the limiting reagent? 2Al + 3CuCl -> 2AlCl3 + 3Cu

The limiting reagent is CuCl

Explanation:

The initial number of moles of aluminum, Al = 4.5 moles

The number of moles of copper, Cu = 6.5 moles

The given chemical reaction is presented as follows;

2Al + 3CuCl → 2AlCl₃ + 3Cu

Therefore, we have, 2 moles of aluminum combine with 3 moles of CuCl produces 2 moles of AlCl₃ and 3 moles of Cu

1 mole of Al will combine with (3/2) moles of CuCl,

Therefore, 4.5 moles of Al will combine with (4.5 × (3/2) = 6.75) 6.75 moles of CuCl

Given that the number of moles of CuCl present is only 6.5 moles which is less than the 6.75 moles required to combine with the 4.5 moles, the 6.5 moles of CuCl limits the amount of AlCl₃ and Cu produced, and therefore the CuCl is the limiting reagent.

"you start with 15.67 g mgco3-xh2o, when heated and all the water is removed, you are left with 7.58 g mgco3 only.

therefore, you had 15.67 - 7.58 = 8.09g h2o  

moles of each :

7.58g mgco3 = ? moles = 7.58g / 83.4g/mole = 0.09moles  

8 09g h2o = ? moles = 8.09g / 18g/mole = 0.44moles  

molar ratio between the two by dividing both numbers by the smallest number :

0.09 / 0.09 = 1  

0.44 / 0.09 = 4.8  

mgco3-5h2o"