Question

Use the reaction below for the decomposition of sodium azide
to sodium metal and nitrogen gas.
2NaN3 (s) → 2Na(s) + 3N2(8)
What volume of nitrogen at STP is generated by the
decomposition of 130.0 g NaN3 ?

Answer 1

Answer:

34.05dm^3 of nitrogen gas

Explanation:

First things first, we need to find the number of moles of Sodium azide. We can do that by using the formula m=nM, mass = no. moles x Molar Mass

Rearrange to solve for no. moles and substituting in the known values and we have:

n = m/M

no. moles = 130.0g / (2x(22.99+3x14.01))

no. moles = 130.0/130.0 (4.s.f.)

no. moles = 1

Now we can use the ratio given in the equation to find the number of moles of Nitrogen that will be made:

1 x 3/2 = 1.5 moles of Nitrogen

Now we use the constant that 1 mole of any gas will always have a volume of 22.7dm^3 at STP.

1.5 x 22.7 = 34.05dm^3 of nitrogen gas.

Hope this helped!