Question #5b51f

1 Answer
Jan 24, 2018

Answer:

Here's one way to do it.

Explanation:

Problem:

Assume you are reacting 20.0 g of carbon and 20.0 g of oxygen to form carbon dioxide. What is the mass of excess reactant used and what mass of excess reactant is unused?

Solution:

#M_r:color(white)(mmmll) 12.01 color(white)(mm)32.00#
#color(white)(mmmmmmm) "C" color(white)(m)+ color(white)(m)"O"_2color(white)(m) →color(white)(m) "CO"_2#
#"Mass/g:"color(white)(mmm)20color(white)(mmml)20#
#"Amt/mol:"color(white)(mll)1.67color(white)(mm)0.625#

You calculated that:

  • The amount of #"O"_2# is 0.625 mol
  • The amount of #"C"# is 1.67 mol
  • #"O"_2# is the limiting reactant
  • #"C"# is the excess reactant

Then

#"Moles of C used" = 0.625 color(red)(cancel(color(black)("mol O"_2))) × "1 mol C"/(1 color(red)(cancel(color(black)("mol O"_2)))) = "0.625 mol C"#

#"Mass of C used" = 0.625 color(red)(cancel(color(black)("mol C"))) × "12.01 g C"/(1 color(red)(cancel(color(black)("mol C")))) = "12.5 g C"#

#"Mass of unused C = 20.0 g - 12.5 g = 7.5 g"#

Therefore,

  • Mass of excess reactant used = 12.5 g
  • Mass of unused excess reactant = 7.5 g