Respuesta :
Answer: The mass of water required will be 10.848 g.
Explanation:
To calculate the number of moles, we use the equation:
[tex]\text{Number of moles}=\frac{\text{Given mass}}{\text{Molar mass}}[/tex] ......(1)
- For nitric acid:
Given mass of nitric acid = 75.9 g
Molar mass of nitric acid = 63 g/mol
Putting values in above equation, we get:
[tex]\text{Moles of nitric acid}=\frac{75.9g}{63g/mol}=1.204mol[/tex]
For the given chemical equation:
[tex]3NO_2(g)+H_2O(l)\rightarrow 2HNO_3(aq.)+NO(g)[/tex]
By Stoichiometry of the reaction:
2 moles of nitric acid is produced by 1 mole of water.
So, 1.204 moles of nitric acid will be produced by = [tex]\frac{1}{2}\times 1.204=0.602mol[/tex] of water.
Now, calculating the amount of water, we use equation 1:
Moles of water = 0.602 mol
Molar mass of water = 18.02 g/mol
Putting values in equation 1, we get:
[tex]0.602=\frac{\text{Mass of water}}{18.02g/mol}\\\\\text{Mass of water}=10.848g[/tex]
Hence, the mass of water required will be 10.848 g.
