Question

What was the volume of N, corrected to STP Online 5. Calculate the moles of N, that should have been produced. 6. What is your experimentally determined molar volume at STP? 7. What is the % error in your experimentally determined molar volume?

Answer 1

The balanced chemical equation for the reaction taking place here is:

HSO₃NH₂ + NaNO₂ = N₂ +NaHSO₄ + H₂O

1. We can see from the above equation that 1 mole of HSO₃NH₂ reacts with 1 mole of NaNO₂ to produce 1 mole of N₂.

Now, molar mass of HSO₃NH₂ is = 97.1 g/mol

molar mass of NaNO₂ = 69 g/mol

Therefore, 97.1 g of HSO₃NH₂ reacts with 69 g of NaNO₂.

According to question, 1.41 g of HSO₃NH₂ is taken which would require = { (69 g/ 97.1 g) * 1.41 g} = 1.002 g of NaNO₂ to react.

Now, from the definition of molarity we know that 1 L of a 1 M solution contains 1 mole of the solute.

Therefore, we have to calculate how much volume 0.80 M solution of NaNO₂ would contain this 1.002 g of NaNO₂.

Now, 1 L of a 1 M solution of NaNO₂ contains 1 mole of NaNO₂ = 69 g of NaNO₂.

1 L of a 0.80 M solution of NaNO₂ contains 0.80 mole of NaNO₂ = (69 g*0.80) = 55.2 g of NaNO₂

Therefore, 1.002 g of NaNO₂ is contained in = (1 L/ 55.2 g)* 1.002 g = 0.018 L of 0.08 M solution of NaNO₂.

Therefore, 0.018 L = 18 mL of the 0.80 M NaNO₂ solution is required.

3. We know from Dalton's law of partial pressures that the total vapor pressure of a mixture of non-reacting gases is equal to the partial pressures of the individual gases.

Therefore, P_total = p_N2 + p_vapor

The vapor pressure (here, p_vapor) of water at 17.5 ^oC is 15.03 torr.

Therefore, partial presssure of N₂, p_N2 = P_total - p_vapor = 782.5 mmHg - 15.03 torr

= 782.5 mmHg - 15.03 mmHg [Since, 1 torr = 1 mmHg]

= 767.47 mmHg

4. We know, P₁V₁/ T₁ = P₂V₂/ T₂

P₁ = pressure at initial condition = 767.47 mmHg

V₁ = volume at initial condition = 352.5 mL

T₁ = temperature at initial condition = 17.5 ^oC = (17.5 + 273) K = 290.5 K

P₂= pressure at final condition = 1 atm = 760 mmHg (STP)

V₂ = volume at initial condition = ? (to be determined)

T₂ = temperature at final condition = 0 ^oC = 273 K (STP)

Therefore, V₂ = P₁V₁T₂/P₂T₁

= 767.47 mmHg * 352.5 mL * 290.5 K / 760 mmHg * 273 K

= 378.78 mL

Therefore, the volume of N₂ corrected to STP conditions is 378.78 mL.

5. We know from the balanced equation that 1 mole of HSO₃NH₂ reacts with 1 mole of NaNO₂ to produce 1 mole of N₂.

Now, 1.41 g of HSO₃NH₂ =(1.41 g/ 97.1 g/mol) =0.0145 moles of HSO₃NH₂

0.0145 moles of HSO₃NH₂ should produce 0.0145 moles of N₂.

Therefore, 0.0145 moles of N₂ should have been produced.

6. From 4 and 5 we know that 0.0145 moles of N₂ takes up 378.78 mL volume at STP conditions.

Therefore, 1 mole of N₂ has the volume =(378.78 mL / 0.0145 mole) = 26122.76 mL.

Therefore, the experimentally determined molar volume of N₂ at STP = 26122.76 mL.

7. From the ideal gas equation, the theoretical value of molar volume (at STP) is = 22.4 L = 22400 mL

Therefore, the % error in the experimentally determined molar volume = { (26122.76 mL - 22400 mL) / 22400mL) } * 100 %

= 16.62 %