how many grams of sodium azide are required to produce 31.0 g of nitrogen
To determine the amount of sodium azide required to produce 31.0 g of nitrogen, we need to use the balanced chemical equation for the reaction:
2 NaN3 → 2 Na + 3 N2
From the equation, we can see that 2 moles of sodium azide (NaN3) produce 3 moles of nitrogen gas (N2).
To find the number of moles of nitrogen gas, we can use the molar mass of nitrogen (N2) which is 28.02 g/mol. We divide the mass of nitrogen given (31.0 g) by the molar mass to get the number of moles:
31.0 g / 28.02 g/mol = 1.107 mol
Since the stoichiometry of the reaction tells us that 2 moles of sodium azide produce 3 moles of nitrogen gas, we can set up a ratio to find the number of moles of sodium azide:
(1.107 mol N2) / (3 mol N2) = (x mol NaN3) / (2 mol NaN3)
Simplifying the ratio, we find:
x mol NaN3 = (1.107 mol N2) * (2 mol NaN3) / (3 mol N2) = 0.738 mol NaN3
Finally, we can convert the number of moles of sodium azide to grams using its molar mass of 65.01 g/mol:
0.738 mol NaN3 * 65.01 g/mol = 47.94 g (rounded to three significant figures)
Therefore, approximately 47.94 grams of sodium azide are required to produce 31.0 g of nitrogen.