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

Jun 4, 2017

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

#### Explanation:

given: $3.6 \text{g } S {b}_{2} {O}_{3}$
reacts with carbon (C)
want: mass of Sb

$S {b}_{2} {O}_{3} + 3 C \rightarrow 2 S b + 3 C O$

Setup a Dimensional analysis starting with the given over 1:

$\frac{3.6 \text{g } S {b}_{2} {O}_{3}}{1}$

Multiply by the molar mass conversion factor for $S {b}_{2} {O}_{3}$:

$\left(3.6 \text{g "Sb_2O_3)/1(1"mol "Sb_2O_3)/(291.5"g } S {b}_{2} {O}_{3}\right)$

From the equation we see that 1 mole of $S {b}_{2} {O}_{3}$ produces 2 moles of $S b$ so we multiply by that conversion factor:

$\left(3.6 \text{g "Sb_2O_3)/1(1"mol "Sb_2O_3)/(291.5"g "Sb_2O_3)(2"mol "Sb)/(1"mol } S {b}_{2} {O}_{3}\right)$

Multiply by the conversion factor for the molar mass of $S b$:

$\left(3.6 \text{g "Sb_2O_3)/1(1"mol "Sb_2O_3)/(291.5"g "Sb_2O_3)(2"mol "Sb)/(1"mol "Sb_2O_3)(121.76"g "Sb)/(1"mol } S b\right)$

Please observe that the units cancel; leaving only the units $\text{g } S b$

(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