How many grams of #O_2# are needed to react with #84.5# g of #NH_3#?

1 Answer
May 24, 2018

Answer:

#"198 g O"_2"# are needed to react with #"84.5 g NH"_3"#.

Explanation:

Balanced equation

#"4NH"_3("g") + "5O"_2("g")"##rarr##"4NO(g) + 6H"_2"O(g)"#

There are three steps required to answer this question.

#color(red)1#. Convert mass #"NH"_3# to mol #"NH"_3"# by dividing the given mass by its molar mass #("17.031 g/mol")#. Do this by multiplying by the reciprocal of the molar mass.

#color(blue)2#. Determine #"mol O"_2# by multiplying #"mol NH"_3"# by the mol ratio between #"NH"_3# and #"O"_2# in the balanced equation, with #"mol O"_2# in the numerator.

#color(green)3#. Determine mass #"O"_2# by multiplying #"mol O"_2"# by its molar mass #("31.998 g/mol")#.

#color(red)(84.5)color(black)cancel(color(red)("g NH"_3))xx(color(red)1color(black)cancel(color(red)("mol NH"_3)))/(color(red)17.031color(black)cancel(color(red)("g NH"_3)))xx(color(blue)5color(black)cancel(color(blue)("mol O"_2)))/(color(blue)4color(black)cancel(color(blue)("mol NH"_3)))xx(color(green)31.998color(black)cancel(color(green)("g O"_2)))/(color(green)1color(black)cancel(color(green)("mol O"_2)))=color(green)("198 g O"_2"# (rounded to three significant figures)

#"198 g O"_2"# are needed to react with #"84.5 g NH"_3"#.