If you had excess chlorine, how many moles of of aluminum chloride could be produced from 10.0 gg of aluminum? Express your answer to three significant figures and include the appropriate units.

The answer is 0.370moles (3 s.f)

Explanation:

Step 1: write the balanced equation for the reaction

Equation for the reaction:

2Al(s) + 6HCl(g) > 2AlCl3(g) + 3H2(g)

Step 2: Equate the mole of the needed substances

So therefore,

2moles of aluminum will produce 2 moles of aluminum chloride

Note that:

Relative atomic mass of Al = 27 g/mole

And, 27g of Al = 1 mole of Al.

Step 3: Solve for the required number of mole.

2moles of Al = 2moles of AlCl3

Same as

1 mole of Al = 1 mole of AlCl3

27g of Al = 1 mole of AlCl3

10g of Al = (1/27 * 10g) of AlCl3

10g of Al = 0.370moles of AlCl3

Thanks