If 1.00 mol of n2 has a volume of 47.0 l under the reaction conditions, how many liters of gas can be formed by heating 38.0 g of nan3? the reaction is: 2nan3→3n2 (g) + 2na

From the periodic table:

mass of sodium = 23 grams

mass of nitrogen = 14 grams

molar mass of NaN3 = 23 + 3 (14) = 65 grams

molar mass of N2 = 2 (14) = 28 grams

From the balanced chemical equation:

2 moles of NaN3 produces 3 moles of N2.

This means that 130 grams of NaN3 produces 84 grams of N2.

To know the amount of N2 produced from 38 grams of NaN3, we will just do a cross multiplication as follows:

amount produced = (38 x 84) / (130) = 24.55 grams

number of moles = mass/molar mass = 24.55 / 28 = 0.876 moles

one mole occupies 47 liters, therefore:

volume = 0.876 x 47 = 41.2153 liters