Question

In the following reaction, how many liters of nitrogen, N2, measured at STP, would be produced from the decomposition of 399 g of sodium azide, NaN3?

2NaN3(s)--------2Na(s)+3N2(g)

Answer 1

The molar mass of NaN3 is 65.03 g/mol

therefore 399 g of NaN3 corresponds to = 399 g / 65.03 (g/mol) = 6.135 mol of NaN3

from the given reaction in the question 2 mol of NaN3 produces 3 mol of N2

therefore 6.135 mol of NaN3 produces (3/2) * 6.135 mol of N2 = 9.203 mol of N2

since 1 mol of any gas occupies 22.41 L at STP

9.203 mol of N2 gas occupies 9.203 * 22.41 L at STP = 206.24 L of N2

therefore 206.24 L of N2 gas is produced from the decomposition of 399g of NaN3 at STP.