Answer:
15.0 liters of HN3 are needed to react completely with 30.0 liters of NO.
Step-by-step explanation:
The amount of HN3 required to react completely with a given volume of NO can be determined by using the balanced chemical equation for the reaction. The reaction between HN3 and NO is given by:
4 HN3 + 5 NO → 4 HNO3 + N2
From the equation, we can see that for every 4 moles of HN3 that react, 5 moles of NO are consumed. To determine the amount of HN3 needed to react with a given volume of NO, we can convert the volume of NO to moles and then use the stoichiometry of the reaction to find the number of moles of HN3 required.
For example, if we have 30.0 liters of NO, we can convert it to moles using the ideal gas law:
30.0 L NO * (1 mole NO / 22.4 L) = 1.34 moles NO
Next, we can use the stoichiometry of the reaction to find the number of moles of HN3 required:
1.34 moles NO / (5 moles NO / 4 moles HN3) = 0.67 moles HN3
Finally, we can convert the number of moles of HN3 to liters:
0.67 moles HN3 * (22.4 L HN3 / 1 mole HN3) = 15.0 L HN3
Therefore, 15.0 liters of HN3 are needed to react completely with 30.0 liters of NO.