How many moles of #N_2# are produced by the decomposition of 1.25 mol of sodium azide in the reaction #2NaN_3(s) -> 2Na(s) + 3N_2(g)#?

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How many moles of #N_2# are produced by the decomposition of 1.25 mol of sodium azide in the reaction #2NaN_3(s) -> 2Na(s) + 3N_2(g)#?

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Answer 1 · 2016-02-28T19:25:37

#"1.25 mol NaN"_3"# will produce #"1.88 mol N"_2"#.

Explanation:

#"2NaN"_3("s")##rarr##"2Na("s")+3N"_2("g")"#

Multiply the given moles of sodium azide by the mole ratio between #"NaN"_3"# and #"N"_2"# from the balanced equation, with #"N"_2"# in the numerator. #"2 mol N"_2:##"3 mol N"_2"#

#1.25cancel"mol NaN"_3xx(3"mol N"_2)/(2cancel"mol NaN"_3)="1.88 mol N"_2"# rounded to three significant figures

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How many moles of #N_2# are produced by the decomposition of 1.25 mol of sodium azide in the reaction #2NaN_3(s) -> 2Na(s) + 3N_2(g)#?

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