Question

If 325 grams of aluminum chloride is reacted, how much magnesium chloride would be produced in this reaction?

Answer 1

First of all, we need to find the moles that are in 325 grams of aluminum chloride. For this, we just divide by the molar mass of the compound which is 133.34g.

Now,

`(325)/(133.34)=2.43\text{molesAlCl}3`

Now, using the coefficients of the reaction:

`2AlCl_3+3MgSO_4\to Al_2(SO_4)+3MgCl_2`
`2.43\text{molAlCl}3\cdot\frac{3molMgCl2}{2\text{molAlCl}3}=3.65molMgCl2`

Finally, the amount in grams can be found multiplying by the molar mass:

`3.25\text{molMgCl}2\cdot\frac{95.21g}{1\text{molMgCl}2}=348gMgCl2`

Therefore, 348g of MgCl2 are produced.