Question

Nitric acid can be prepared by bubbling dinitrogen pent- Oxide into water N2O5(g)H20(e)2 H*(aq) 2 NO (aq) (a) How many moles of H are obtained when 1.50 L of N2O5 at 25°C and 1.00 atm pressure is bubbled into water? (b) The solution obtained in (a) after reaction is com- plete has a volume of 437 mL. What is the molarity of the nitric acid obtained?

Answer 1

Answer :

(a) : From the reaction , we can see that 1 mole of N_{​​​​​​2} O_{​​​​​​5} produced 2 mole of H⁺ .

So, first step is to calculate moles of N_{​​​​​​2} O_{​​​​​​5} gas using ideal gas equation :

P * V = n * R * T

n ( no. Of moles ) = P * V / (R * T )

R (gas constant) = 0.0821 L atm mol^{​​​​-1} K^{​​​​​​-1}

T ( temperature ) = 25°C = 298 K

V ( volume) = 1.5 L

P (pressure ) = 1 atm

Put all these values to calculate no. of moles

n = 1 * 1.5 / ( 0.0821 * 298 ) = 0.0613 moles

So, moles of N_{​​​​​​2} O_{​​​​​​5} used = 0.0613 moles

So, moles of H⁺ obtained = 2 * moles of N_{​​​​​​2} O_{​​​​​​5} used = 2 * 0.0613 = 0.1226 moles .

So, moles of H⁺ obtained = 0.1226 moles .

(b) : To calculate Molarity of nitric acid obtained

Moles of nitric acid ( HNO_{​​​​3}​​) obtained after reaction = 0.1226

Volume of solution = 437 ml = 0.437 L

Molarity of nitric acid solution = moles of nitric acid obtained / volume of solution (L) = 0.1226 / 0.437 = 0.28 M

So, Molarity of nitric acid obtained = 0.28 M