If 15.0 g of NaN3 is in a air bag, how much Nitrogen gas can be produced.
how many moles of NaN3 in 15g?
Use the reaction equation to see how many moles of N2 will be produced.
Each mole occupies 22.4L at STP.
The reaction equation i was given is...
2NaN3 ---> 2Na + 3N2
To determine how much nitrogen gas can be produced from 15.0 g of NaN3, we need to consider the stoichiometry of the reaction. The balanced chemical equation for the decomposition reaction of Sodium Azide (NaN3) is:
2 NaN3(s) -> 2 Na(s) + 3 N2(g)
According to the equation, 2 moles of NaN3 will produce 3 moles of N2 gas.
First, we need to calculate the molar mass of NaN3.
The molar mass of sodium (Na) is 22.99 g/mol, and the molar mass of nitrogen (N) is 14.01 g/mol. Therefore, the molar mass of NaN3 can be calculated as:
Molar mass of NaN3 = (22.99 g/mol) + 3(14.01 g/mol) = 65.00 g/mol
Next, we need to convert the mass of NaN3 (15.0 g) to moles.
Moles of NaN3 = mass of NaN3 / molar mass of NaN3
= 15.0 g / 65.00 g/mol
Now, we can calculate the moles of N2 gas produced using the stoichiometric ratio.
Moles of N2 = Moles of NaN3 x (3 moles of N2 / 2 moles of NaN3)
Finally, we can convert the moles of N2 gas to grams.
Mass of N2 = Moles of N2 x molar mass of N2
= Moles of N2 x (2 x 14.01 g/mol)
By following these steps, you can calculate how much nitrogen gas can be produced from 15.0 g of NaN3.