How many grams of Na(l) are produced per litre of N2(g) formed in the decomposition of sodium azide, NaN3 if the gas is collected at 25 degrees celsius and 1.0 bar (.98692 atm)? The R constant is 0.08206 Latm/Kmol

1 Answer
Mar 27, 2016

Answer:

The reaction produces #"0.618 g Na"//"L N"_2#.

Explanation:

We aren’t given the amount of #"NaN_3"#, so let's assume that we have 1 mol of #"NaN"_3#.

There are six steps involved in this problem:

  1. Write the balanced equation for the reaction.
  2. Use the molar ratio of #"Na":"NaN"_3# from the balanced equation to calculate the moles of #"Na"#.
  3. Use the molar mass of #"Na"# to calculate the mass of #"Na"#.
  4. Use the molar ratio of #"N"_2:"NaN"_3# from the balanced equation to calculate the moles of #"N"_2#.
  5. Use the Ideal Gas Law to calculate the volume of #"N"_2#.
  6. Calculate the mass of #"Na"# per litre of #"N"_2#.

Let's get started.

Step 1. Write the balanced chemical equation.

#"2NaN"_3 → "2Na" + "3N"_2#

2. Calculate the moles of #"Na"#

#"moles of Na" = 1 color(red)(cancel(color(black)("mol NaN"_3))) × "2 mol Na"/(2 color(red)(cancel(color(black)("mol NaN"_3)))) = "1 mol Na"#

3. Calculate the mass of #"Na"#.

#"mass of Na" = 1 color(red)(cancel(color(black)("mol Na"))) × "22.99 g Na"/(1 color(red)(cancel(color(black)("mol Na")))) = "22.99 g Na"#

4. Calculate the moles of #"N"_2#.

#"moles of N"_2 = 1 color(red)(cancel(color(black)("mol NaN"_3))) × ("3 mol N"_2)/(2 color(red)(cancel(color(black)("mol NaN"_3)))) = "1.5 mol N"_2#

5. Calculate the volume of #"N"_2#.

The Ideal Gas Law is

#color(blue)(|bar(ul(PV = nRT)|)#

#P = "1 bar" = "0.986 92 atm"#
#n= "1.5 mol"#
#R = "0.082 06 L·atm·K"^"-1""mol"^"-1"#
#T = "25 °C = 298.15 K"#

We can rearrange the Ideal Gas Law to get:

#V = (nRT)/P = (1.5 color(red)(cancel(color(black)("mol"))) × "0.082 06 L·"color(red)(cancel(color(black)("atm·K"^"-1""mol"^"-1"))) × 298.15 color(red)(cancel(color(black)("K"))))/("0.986 92" color(red)(cancel(color(black)("atm")))) = "37.19 L"#

5. Calculate the #"Na:N"_2# ratio.

#"Na"/"N"_2 = "22.99 g Na"/("37.19 L N"_2) = "0.618 g Na"//"L N"_2#