The chemical reaction for the generation of gas in an automobile safety air bag is 2NaN3(s)→ 2Na(s)+3N2(g)
What volume of gas is produced if there are 130.0 grams of NaN3 used in the
reaction? (The density of nitrogen gas is 0.916 g/L.) Please show your work.
i need help
130 gm is 2 mol of NaN3 so we will get 3 mol of N2 gas as output
That corresponds to 84 gms of mass which corresponds to 91.7 L of Nitrogen at some elevated temperature (i.e above STP)
To find the volume of gas produced in the reaction, you need to use the given information and the molar mass of NaN3.
First, determine the number of moles of NaN3:
Molar mass of NaN3 = (22.99 g/mol + 14.01 g/mol + (3 * 16.00 g/mol)) ≈ 65.01 g/mol
Number of moles of NaN3 = mass of NaN3 / molar mass of NaN3
= 130.0 g / 65.01 g/mol
≈ 1.998 mol
From the balanced chemical equation, you can see that for every 2 moles of NaN3, 3 moles of N2 are produced. Therefore, you have:
Number of moles of N2 = (3 mol N2 / 2 mol NaN3) * (1.998 mol NaN3)
≈ 2.997 mol N2
Next, you can calculate the volume of N2 gas using the given density:
Volume of N2 = mass of N2 / density of N2
= (2.997 mol N2) * (28.02 g/mol) / (0.916 g/L)
≈ 92.33 L
Therefore, the volume of gas produced in the reaction is approximately 92.33 liters.
2NaN3 ==> 2Na + 3N2
mols N2 gas = grams/molar mass = x mols N2
Using the coefficients in the balanced equation, convert mols N2 to mols NaN3. That's x mols N2*(2 mols NaN3/3 mols N2) = x mols N2*(2/3) = ? mols NaN3.
Then ? mols NaN3 x molar mass NaN3 = grams NaN3.