In the chemical reaction #"C"_6"H"_6"OS" + "8O"_2"##rarr##"6CO"_2 + "3H"_2"O" + "SO"_2#, how many moles of #"C"_6"H"_6"OS"# are needed to react with #"317 g O"_2"#?

1 Answer
Meave60
Jul 12, 2017

#"1.24 mol C"_6"H"_6"OS"# are needed to react with #"317 g O"_2#.

Explanation:

Balanced equation:

#"C"_6"H"_6"OS" + "8O"_2"##rarr##"6CO"_2 + "3H"_2"O" + "SO"_2#

First convert the given mass of #"O"_2# to moles. The molar mass of #"O"_2# is #"31.998 g/mol"#.

#317color(red)cancel(color(black)("g O"_2))xx(1"mol O"_2)/(31.998color(red)cancel(color(black)("g O"_2)))="9.91 mol O"_2#

Determine the moles #"C"_6"H"_6"OS"# that will react with #"317 g O"_2"# by multiplying moles of #"O"_2# by the mole ratio between #"C"_6"H"_6"OS"# and #"O"_2# from the balanced equation so that moles #"O"_2# are canceled.

#9.91color(red)cancel(color(black)("mol O"_2))xx(1"mol C"_6"H"_6"OS")/(8color(red)cancel(color(black)("mol O"_2)))="1.24 mol C"_6"H"_6"OS"# (rounded to three sig figs)

Related questions
See all questions in Stoichiometry
Impact of this question
1648 views around the world
You can reuse this answer
Creative Commons License