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16 January, 08:34

Use this equation for the following problems: 2NaN3 - -> 2Na+3N2

density of N2=0.92g/L

How many grams of NaN3 are needed to make 23.6L of N2?

How many mL of N2 result if 8.3g Na are also produced?

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Answers (1)
  1. 16 January, 09:02
    0
    1) 65.0

    2) 16.434 L = 16434 mL.

    Explanation:

    2NaN₃ → 2Na + 3N₂,

    It is clear from the balanced equation that 2.0 moles of NaN₃ are decomposed to 2.0 moles of Na and 3.0 moles of N₂.

    Q1: How many grams of NaN₃ are needed to make 23.6L of N₂?

    Density of N₂ = 0.92 g/L which means that every 1.0 L of N₂ contains 0.92 g of N₂.

    Now, we can get the mass of N₂ in 23.6 L N₂ using cross multiplication:

    1.0 L of N₂ contains → 0.92 g of N₂.

    23.6 L of N₂ contains →? g of N₂.

    ∴ The mass of N₂ in 23.6 L of N₂ = (23.6 L) (0.92 g) / (1.0 L) = 21.712 g.

    We can get the no. of moles of 23.6 L of N₂ (21.712 g) using the relation:

    n = mass/molar mass = (21.712 g) / (28.0 g/mol) = 0.775 mol.

    We can get the no. of moles of NaN₃ needed to produce 0.775 mol of N₂:

    using cross multiplication:

    2.0 moles of NaN₃ produce → 3.0 moles of N₂, from the balanced equation.

    ? mol of NaN₃ produce → 0.775 moles of N₂.

    ∴ The no. of moles of NaN₃ needed = (2.0 mol) (0.775 mol) / (3.0 mol) = 0.517 mol.

    Finally, we can get the grams of NaN₃ needed:

    mass = no. of moles x molar mass = (0.517 mol) (65.0 g/mol) = 33.6 g.

    Q2: How many mL of N₂ result if 8.3 g Na are also produced?

    We need to get the no. of moles of 8.3 g Na using the relation:

    n = mass/atomic mass = (8.3 g) / (22.98 g/mol) = 0.36 mol.

    We can get the no. of moles of N₂ produced with 0.36 mol of Na:

    using cross multiplication:

    2.0 moles of Na produced with → 3.0 moles of N₂, from the balanced equation.

    0.36 moles of Na produced with →? moles of N₂.

    ∴ The no. of moles of N₂ needed = (3.0 mol) (0.36 mol) / (2.0 mol) = 0.54 mol.

    We can get the mass of 0.54 mol of N₂:

    mass = no. of moles x molar mass = (0.54 mol) (28.0 g/mol) = 15.12 g.

    Now, we can get the mL of 15.12 g of N₂:

    using cross multiplication:

    1.0 L of N₂ contains → 0.92 g of N₂, from density of N₂ = 0.92 g/L.

    ? L of N₂ contains → 15.12 g of N₂.

    ∴ The volume of N₂ result = (1.0 L) (15.12 g) / (0.92 g) = 16.434 L = 16434 mL.
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