One in a series of reactions that inflate air bags in automobiles is the decomposition of sodium azide (NaN3).
2 NaN3(s) 2 Na(s) + 3 N2(g)
Determine the mass of N2 produced if 279.8 g NaN3 is decomposed.
What would be the grams?
Convert g NaN3 to mols. mols = g/molar mass.
Convert mols NaN3 to mols N2 using the coefficients in the balanced equation.
Convert mols N2 to grams using g = mols x molar mass.
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To determine the mass of N2 produced, we will use the molar mass of NaN3 and the stoichiometry of the balanced equation.
The molar mass of NaN3 can be calculated as follows:
Na: 1 atom x 22.99 g/mol = 22.99 g/mol
N: 1 atom x 14.01 g/mol = 14.01 g/mol
3 atoms of N: 3 atoms x 14.01 g/mol = 42.03 g/mol
So, the molar mass of NaN3 is 65.04 g/mol.
Now, we can set up a ratio using the balanced equation:
2 mol NaN3 : 3 mol N2
65.04 g NaN3 : x g N2
To find the grams of N2 produced, we can use the following calculation:
279.8 g NaN3 x (3 mol N2 / 2 mol NaN3) x (65.04 g N2 / 1 mol N2) = 270.6 g N2
Therefore, the mass of N2 produced when 279.8 g of NaN3 is decomposed is 270.6 grams.
To determine the mass of N2 produced when 279.8 g of NaN3 is decomposed, we need to use stoichiometry. Stoichiometry is a quantitative relationship between reactants and products of a chemical reaction.
First, we need to calculate the molar mass of NaN3:
Na: atomic mass = 22.99 g/mol
N: atomic mass = 14.01 g/mol
3 × N: (3 × 14.01) = 42.03 g/mol
The molar mass of NaN3 = 22.99 + 42.03 = 65.02 g/mol
Next, we need to convert the given mass of NaN3 to moles:
Moles of NaN3 = mass(g) / molar mass(g/mol)
Moles of NaN3 = 279.8 g / 65.02 g/mol ≈ 4.30 mol
According to the balanced equation, 2 moles of NaN3 produces 3 moles of N2. Therefore, 4.30 moles of NaN3 will produce:
(3 moles of N2 / 2 moles of NaN3) × 4.30 moles of NaN3 = 6.45 moles of N2
Lastly, we convert the moles of N2 to grams using the molar mass of N2:
Molar mass of N2 = 2 × 14.01 g/mol = 28.02 g/mol
Mass of N2 = moles × molar mass = 6.45 mol × 28.02 g/mol ≈ 180.39 g
Therefore, the mass of N2 produced when 279.8 g of NaN3 is decomposed is approximately 180.39 grams.