how many grams of sodium azide are required to form 16.55g of nitrogen gas in 2 N a N 3 ( s

Question

how many grams of sodium azide are required to form 16.55g of nitrogen gas in
2
N
a
N
3
(
s
)

2
N
a
(
s
)
+
3
N
2
(
g
)
?

in progress 0
Khoii Minh 1 day 2021-07-21T04:58:14+00:00 1 Answers 0 views 0

Answers ( )

    0
    2021-07-21T05:00:07+00:00

    Answer:

    16.8128 grams of sodium azide is needed

    Explanation:

    2NaN3 = 2Na + 3N2

    molar mass of NaN3 =

    2(11 + (7  \times 3) = 64

    molar mass of N2

    3(7 \times 2) = 42

    moles of N2 =

    moles=mass/R.F.M

    16.55÷42 = 0.3940

    mole ratio=

    2NaN3 : 3N2

    3 = 0.3940

    2=?

    0.394. \times 2 \div 3 = 0.2627

    Mass of sodium azide =64×0.2627

    =16.8126

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Giải phương trình 1 ẩn: x + 2 - 2(x + 1) = -x . Hỏi x = ? ( )