Calculate the mass of nitric oxide that will just neutralize dilute nitric acid containing 12.6g of nitric (v) acid in one litre of water
To calculate the mass of nitric oxide that will neutralize the dilute nitric acid, we need to determine the balanced equation for the reaction between nitric oxide and nitric acid and calculate the molar ratio.
The balanced equation for the reaction between nitric oxide (NO) and nitric acid (HNO3) is:
2NO + HNO3 → 3NO2 + H2O
From the balanced equation, we can see that 2 moles of nitric oxide react with 1 mole of nitric acid, which means the molar ratio is 2:1.
To find the molar mass of nitric oxide (NO), we sum the atomic masses of nitrogen (N) and oxygen (O):
Molar mass of NO = atomic mass of N + atomic mass of O
= 14.01 g/mol + 16.00 g/mol
= 30.01 g/mol
Now we can calculate the moles of nitric acid (HNO3) using its given mass of 12.6g and molar mass:
Moles of HNO3 = mass of HNO3 / molar mass of HNO3
= 12.6 g / (1 mol HNO3 / 63.01 g HNO3)
= 0.20 mol HNO3
Since the molar ratio of NO to HNO3 is 2:1, the number of moles of NO needed to react with 0.20 mol of HNO3 would be half of this:
Moles of NO = 0.20 mol HNO3 / 2
= 0.10 mol NO
Finally, we can calculate the mass of NO using its molar mass and the number of moles:
Mass of NO = moles of NO * molar mass of NO
= 0.10 mol * 30.01 g/mol
= 3.00g
Therefore, the mass of nitric oxide that will just neutralize the dilute nitric acid is 3.00g.