The number of moles of sodium nitrate, NaNO3, required to produce .056m mole of oxygen. (NaNO2 isthe other product.)?

1 Answer
Meave60
Mar 19, 2018

You need #"1.1 mol NaNO"_3"# to produce #"0.56 mol O"_2"#.

Explanation:

I am going to answer based on #"0.56 mol O"_2"#.

Balanced equation

#"2NaNO"_3("s")stackrel(Delta" ")(->)"2NaNO"_2("s") + "O"_2("g")"#

Multiply the given mol #"O"_2"# by the mole ratio between #"NaNO"_3"# and #"O"_2"# from the balanced equation, with mol #"NaNO"_3"# in the numerator.

#0.56color(red)cancel(color(black)("mol O"_2))xx(2"mol NaNO"_3)/(1color(red)cancel(color(black)("mol O"_2)))="1.12 mol NaNO"_3"##rArr##"1.1 mol NaNO"_3"# rounded to two significant figures.

You can also do this mentally by looking at the mol ratio between #"NaNO"_3"# and #"O"_2"#, which is #2:1#. So, there will always be twice as many moles #"NaNO"_3"# than moles #"O"_2"#.

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