Question #aba1e

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Question #aba1e

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Answer 1 · 2017-02-02T13:12:02

Balanced equation of the reaction

$N H_{4} N O_{3} \to N_{2} O (g) + 2 H_{2} O$ $NH_4NO_3" "->" "N_2O(g)+2H_2O$

$1 mol 1 mol$ $" "1"mol"" "" "" "" "1"mol"$

$=80g =22400mL at STP$ $" ""=80g"" "" "" "" ""=22400mL"" ""at STP"$

$x g=? 250mL at STP$ $" ""x" "g=?"" "" "" "" "" 250mL"" ""at STP"$

So

$\frac{x}{80} = \frac{250}{22400}$ $x/80=250/22400$

$\Rightarrow x = \frac{80 \times 250}{22400} g \approx 0.89 g$ $=>x=(80xx250)/22400g~~0.89g$

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