Question

How would I determine how many grams of antimony would form from the reaction of carbon with `"3.6 g"` of antimony(III) oxide?

Answer 1

Use dimensional analysis (a.k.a. the factor-label method)

Explanation:

given: `3.6"g "Sb_2O_3`
reacts with carbon (C)
want: mass of Sb

`Sb_2O_3 + 3C rarr 2Sb + 3CO`

Setup a Dimensional analysis starting with the given over 1:

`(3.6"g "Sb_2O_3)/1`

Multiply by the molar mass conversion factor for `Sb_2O_3`:

`(3.6"g "Sb_2O_3)/1(1"mol "Sb_2O_3)/(291.5"g "Sb_2O_3)`

From the equation we see that 1 mole of `Sb_2O_3` produces 2 moles of `Sb` so we multiply by that conversion factor:

`(3.6"g "Sb_2O_3)/1(1"mol "Sb_2O_3)/(291.5"g "Sb_2O_3)(2"mol "Sb)/(1"mol "Sb_2O_3)`

Multiply by the conversion factor for the molar mass of `Sb`:

`(3.6"g "Sb_2O_3)/1(1"mol "Sb_2O_3)/(291.5"g "Sb_2O_3)(2"mol "Sb)/(1"mol "Sb_2O_3)(121.76"g "Sb)/(1"mol "Sb)`

Please observe that the units cancel; leaving only the units `"g "Sb`

`(3.6"g "Sb_2O_3)/1(1"mol "Sb_2O_3)/(291.5"g "Sb_2O_3)(2"mol "Sb)/(1"mol "Sb_2O_3)(121.76"g "Sb)/(1"mol" Sb) = 3.0"g "Sb larr`
answer