calculate the volume of Nitrogen gas to produce 5.6 dm cube of nitric oxide at STP ,N2 + O2 = 2NO
N2 + O2 ==> 2NO
5.6 dm^3 NO x (1 mol N2/2 mol NO) = 5.6 x 1/2 = ?
To calculate the volume of nitrogen gas required to produce a certain volume of nitric oxide at STP (Standard Temperature and Pressure), we first need to understand the molar ratio of nitrogen gas to nitric oxide in the balanced chemical equation.
From the balanced equation: N2 + O2 → 2NO
We can see that it takes 1 mole of N2 to produce 2 moles of NO.
Now, we need to determine the volume of nitric oxide produced. You mentioned that the volume of nitric oxide is 5.6 dm³.
At STP, 1 mole of any gas occupies 22.4 dm³. This is known as molar volume.
So, to calculate the volume of nitrogen gas required, we follow these steps:
Step 1: Calculate the number of moles of nitric oxide using the volume.
Number of moles of NO = Volume of NO / Molar volume
Number of moles of NO = 5.6 dm³ / 22.4 dm³/mol
(Number of moles of NO = 0.25 mol)
Step 2: Since the molar ratio between nitrogen gas and nitric oxide is 1:2, we need twice the number of moles of nitrogen gas.
Number of moles of N2 = 2 * Number of moles of NO
Number of moles of N2 = 2 * 0.25 mol
(Number of moles of N2 = 0.5 mol)
Step 3: Finally, calculate the volume of nitrogen gas using the number of moles and the molar volume.
Volume of N2 = Number of moles of N2 * Molar volume
Volume of N2 = 0.5 mol * 22.4 dm³/mol
(Volume of N2 = 11.2 dm³)
Therefore, the volume of nitrogen gas required to produce 5.6 dm³ of nitric oxide at STP is 11.2 dm³.