Question #a5357

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Question #a5357

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Answer 1 · 2015-08-14T10:00:52

You would need 130 g of sodium azide.

Explanation:

Start with the balanced chemical equation for this decomposition reaction

$2 {NaN}_{3 (s)} \to 2 {Na}_{(s]} + 3 N_{2(g]}$ $color(red)(2)"NaN"_(3(s)) -> 2"Na"_text((s]) + color(blue)(3)"N"_text(2(g])$

Notice that you have a $2 : 3$ $color(red)(2):color(blue)(3)$ mole ratio between sodium azide and nitrogen gas. This ratio tells you how many moles of nitrogen are produced when a certain number of moles of sodium azide reacts.

SImply pout, regardless of how many moles of sodium azide undergo decomposition, your reaction will produce $\frac{3}{2}$ $3/2$ times more moles of nitrogen gas.

In your case, you have to work backward. You know that the reaction produced enough moles to occupy 75 L at those conditions of pressure and temperature.

Use the ideal gas law equation to determine how many moles of nitrogen were produced

$P V = n R T \Rightarrow n = \frac{P V}{R T}$ $PV = nRT implies n = (PV)/(RT)$

$n_{N_{2}} = \frac{\frac{99707}{101325} atm \cdot 75 L}{0.082 \frac{atm \cdot L}{mol \cdot K} \cdot (273.15 + 25) K} = 3.02 moles N_{2}$ $n_(N_2) = (99707/101325color(red)(cancel(color(black)("atm"))) * 75color(red)(cancel(color(black)("L"))))/(0.082(color(red)(cancel(color(black)("atm"))) * color(red)(cancel(color(black)("L"))))/("mol" * color(red)(cancel(color(black)("K")))) * (273.15 + 25)color(red)(cancel(color(black)("K")))) = "3.02 moles N"""_2$

This means that you must have started with

$3.02 {moles N}_{2} \cdot \frac{2 moles NaN_{3}}{3 {moles N}_{2}} = 2.013 moles NaN_{3}$ $3.02color(red)(cancel(color(black)("moles N"_2))) * (color(red)(2)" moles NaN"""_3)/(color(blue)(3)color(red)(cancel(color(black)("moles N"_2)))) = "2.013 moles NaN"""_3$

Now use sodium azide's molar amss to see how many grams would contain this many moles

$2.013 moles \cdot \frac{65.01 g}{1 mole} = 130.87 g$ $2.013color(red)(cancel(color(black)("moles"))) * "65.01 g"/(1color(red)(cancel(color(black)("mole")))) = "130.87 g"$

Rounded to two sig figs, the number of sig figs you gave for the temperature and volume of the gas, the answer will be

$m_{N a N_{3}} = 130 g$ $m_(NaN_3) = color(green)("130 g")$

A cool video on sodium azide's decomposition

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Question #a5357

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