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How many grams of aluminum chloride must decompose to produce 78.3 milliliters of aluminum metal, if the density of aluminum is 2.70 g/mL? Show all steps of your calculation as well as the final answer.
AlCl3 → Al + Cl2
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First find the mass of aluminium required using
m=mass=volume * density
Pay attention to units and significant figures in the calculations.
Next, using the given equation:
AlCl3 → Al + Cl2
Find the molecular mass of AlCl3 (m1) and Al (m2).
Use proportions to find x=actual mass of AlCl3as follows:
solve for x.
Step one balance your chemical reaction:
To do this you need to ensure that the number of atoms is the same on both sides of the equation.
2AlCl3 ----> 2Al + 3Cl2
Conservation of matter. Matter can neither be created nor destroyed.
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You know that the density is 2.70g/mL of aluminum.
So you have 78.3mL produced x (2.70g/mL)
This is dimensional analysis, so when you do this the number of mL cancels out because it's in the denominator: mLx(1/mL) will give you grams as your resulting unit so it would be 211.3g of aluminum produced.
This was the number of grams of aluminum produced by your reaction. What you must do now is to convert this to moles.
211.41g Al x (1mol Al/27g) = 7.8 mol of aluminum.
Now you multiply by the molar Ratio of aluminum to aluminum tri chloride which is
(2Alcl3/2Al) pay attention to the fact that when you multiply these out you'll end up with AlCl3 in the numerator so that's
7.8 mol of AlCl3
Now all you need to do is find the number of grams of AlCl3 that reacted which is quite simple now that you know you had 7.8 moles that reacted.
Look up dimensional analysis because that will help you solve all of these kinds of problems. It's fundamental to any problems involving calculations with moles and conversions.