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"1.1 mol NaNO3 to produce "0.56 mol O"_2"0.56 mol O2.

Explanation:

I am going to answer based on "0.56 mol O"_2"0.56 mol O2.

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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