Question

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?

Answer 1

use the density of aluminum as a conversion factor (or fraction)

Answer 2

you need to do this in a multi-step process. You need to start with the volume of aluminum you need to produce, then find the mass of aluminum that creates it.

you need to balance the reaction first. Can you do that?

Answer 3

\(\_AlCl_3 \rightarrow \_Al + \_Cl_2\)

Answer 4

it's important that you use a reaction arrow, not an equals sign. chemical reactions are not the same thing as algebraic equations

Answer 5

you need to produce 78.3mL of aluminum, but the coefficients of a balanced reaction don't use mL or even grams, they use MOLES, right?

Answer 6

you need to take the mL of aluminum the problem gives you, and turn it into MOLES

you need to use a conversion fraction to do that

Answer 7

the density of aluminum is a great fraction, because it has both MASS and VOLUME in its unit (g/mL)

Answer 8

\[78.3\cancel{mL \space Al} * (\frac{2.7g \space Al}{1\cancel{mL \space Al}})\]

Answer 9

cancelling the UNITS in a problem will always work better than trying to remember whether or not to multiply or divide

Answer 10

yes, and what UNIT does it have?

Answer 11

exactly, GJ

Answer 12

that's CLOSER to moles, but we can't use the equation yet

Answer 13

do you have a relationship between the grams of a substance and its moles?

Answer 14

the relationship we use as the conversion fraction is the MOLAR MASS

Answer 15

yes

Answer 16

i'm going to use 27 just because i'm a little lazy, but see how the molar mass has 2 units in it, and it's in a fraction?

Answer 17

\[211.41\cancel{g Al} * (\frac{1mol \space Al}{27\cancel{g \space Al}})\]

Answer 18

good. NOW we can use the balanced reaction.

Answer 19

See how 2 moles of \(Al\) are produced every time 2 moles of \(AlCl_3\) are used?

Answer 20

the coefficients in the balanced reaction act just like another conversion fraction

Answer 21

\[7.83\cancel{mol \space Al} * (\frac{2mol \space AlCl_3}{2\cancel{mol \space Al}})\]

Answer 22

you divided by 2, but you have to multiply by 2 again, look at the fraction, it's 2/2

Answer 23

it is, but now it has units of MOLES of \(AlCl_3\), not moles of \(Al\)

Answer 24

almost. there's 1 more step

Answer 25

do you see how we used the molar mass to go from grams of Al to moles of Al?

Answer 26

well we're going to do the same thing, but upside down, to go from moles back to grams

Answer 27

but we need the molar mass of \(AlCl_3\)

Answer 28

good. now use that as the last fraction

Answer 29

\[7.83\cancel{mol \space AlCl_3}* (\frac{133.3g \space AlCl_3}{1\cancel{mol \space AlCl_3}})\]

Answer 30

that's it

Answer 31

YW. stoichiometry problems are basically all the same. Find the moles of what you start with, use the coefficients from the balanced reaction to convert, and then find the grams of the "other"