At STP how many L of N2 are produced if 1 mol of NaN3 and .200 moles of KNO3 are in the air bag?
To calculate the volume of N2 produced at STP, you need to determine the limiting reactant by comparing the stoichiometry of the balanced chemical equation and the moles of each reactant.
The balanced chemical equation for the decomposition of NaN3 is as follows:
2 NaN3 --> 2 Na + 3 N2
From the equation, you can see that for every 2 moles of NaN3 used, 3 moles of N2 are produced.
Given:
- Moles of NaN3 = 1 mol
- Moles of KNO3 = 0.200 mol
First, we need to calculate the moles of N2 produced from NaN3:
1 mol NaN3 * (3 mol N2 / 2 mol NaN3) = 1.5 mol N2
Next, we compare the moles of N2 produced from NaN3 and KNO3. Since the moles of N2 produced from NaN3 (1.5 mol) are higher than the moles of N2 that can be produced from KNO3 (0.200 mol), we conclude that KNO3 is the limiting reactant.
Now, we calculate the moles of N2 produced from KNO3:
0.200 mol KNO3 * (3 mol N2 / 2 mol KNO3) = 0.300 mol N2
Finally, using the ideal gas law at STP (Standard Temperature and Pressure), where 1 mole of any ideal gas occupies 22.4 liters, we can convert the moles of N2 to liters:
0.300 mol N2 * (22.4 L / 1 mol) = 6.72 L N2
Therefore, at STP, 0.200 moles of KNO3 will produce approximately 6.72 liters of N2 gas.