Answer to Question #119108 in Chemistry for Aaliyah Michael

Question

How much energy (in joules) will be given off from the following reaction, if 3.0g of sodium reacts with 1.2g of nitrogen in the following reaction? 2Na+ N2 ——> 2NaN3 delta H= -42.7kJ

Answer 1

In the reaction

2Na + 3N₂ $\rightarrow$ 2NaN₃,

2 moles of sodium react with 3 moles of nitrogen:

$n(Na)/2 = n(N_2)/3$ .

The number of the moles contained in 3 g of sodium is its mass divided by its molar mass $M$ (22.99 g/mol):

$n(Na) = \frac{m(Na)}{M(Na)} = \frac{3.0}{22.99} = 0.1305$ mol.

The number of the moles of nitrogen can also be calculated from its mass ($M_{N_2} =$ 28.01 g/mol):

$n(N_2) = \frac{m(N_2)}{M(N_2)} = \frac{1.2}{28.01} = 0.04284$ mol.

Let's compare the equivalent number of the moles of sodium and of nitrogen :

$3·n(Na)= 0.3915; 2·n(N_2) = 0.08568 \Rightarrow 3n(Na)>2n(N_2)$ .

Therefore, there is an excess of sodium. It means that in order to calculate the energy released in the system, we must use the number of the moles of nitrogen. As 3 moles of nitrogen are used in the reaction, 3 moles of nitrogen are equivalent to the release of -42.7 kJ. So, the energy released is the product of the molar enthalpy and the number of the moles of nitrogen, divided by the conversion factor for nitrogen ( 3 mol N₂) :

$Q = ∆H·n(N_2)/3 = 42.7·0.04284/3 = -0.610$ kJ.

Answer: -0.610 kJ.