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

Because electrons have negligible mass, to account for the mass of the isotope, there must be 16 neutrons, 16 neutrally charged, massive, fundamental particles present in the sulfur nucleus. and thus finally, if we have the 32s2− ion, there are 18 electrons, 16 protons, and 16 neutrons.