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

Question

Noah Aldrich · Accepted Answer

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

given: #3.6"g "Sb_2O_3# reacts with carbon (C) want: mass of Sb #Sb_2O_3 + 3C rarr 2Sb + 3CO# Establish a Dimensional analysis using the above 1 as a starting point: #(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

Jose Ayala · Answer

undefined To determine the number of grams of antimony formed from the reaction of carbon with "3.6 g" of antimony(III) oxide, you need to balance the chemical equation for the reaction, calculate the molar mass of antimony(III) oxide, and then use stoichiometry to find the mass of antimony formed. The balanced chemical equation for the reaction is:

2Sb2O3 + 3C → 4Sb + 3CO2

First, calculate the molar mass of antimony(III) oxide (Sb2O3):
Sb: 2 × 121.76 g/mol = 243.52 g/mol
O: 3 × 16.00 g/mol = 48.00 g/mol
Total molar mass = 243.52 g/mol + 48.00 g/mol = 291.52 g/mol

Now, use stoichiometry to find the mass of antimony formed:
1 mole of Sb2O3 produces 4 moles of Sb (from the balanced equation).
So, the molar mass of antimony is 121.76 g/mol.

Using the molar masses, you can set up a ratio:
(121.76 g Sb / 2 × 291.52 g Sb2O3) = (x g Sb / 3.6 g Sb2O3)

Solving for x (the mass of antimony formed):
x = (3.6 g Sb2O3 × 121.76 g Sb) / (2 × 291.52 g Sb2O3)
x ≈ 0.609 g Sb

Therefore, approximately 0.609 grams of antimony would form from the reaction.