HSO 3 NH 2 + NaNO 2 ------> N 2 + NaHSO 4 + H 2 O a. When N 2 was collected over

Question

HSO3NH2 + NaNO2 ------> N2 + NaHSO4 + H2O

a. When N2 was collected over water at 20°C, the totalt pressure was known to equal atmospheric pressure, which was 755.5 torr. Calculate the partial pressure of the nitrogen gas.

b. Calculate the moles of nitrogen present with 425ml of N2 are collected over water at 20°C and 755.5 torr.

c. How many moles of HSO3NH2 would have to react to produce the number of moles of nitrogen calculated in problem b? How many grams of HSO3NH4?

d. Assume that the HSO3NH2 sample weighed 3.125g. Calculate the percent purity of the sulfamic acid.

Explanation / Answer

Solution :-

a) Total pressure is the partial pressure of the N2 + vapor pressure of the H2O

At 20 C vapor pressure of the water is 17.54 torr

Therefore using this value we can calculate the partial pressure of the N2

Partial pressure of N2 = total pressure – vapor pressure of water

                                      = 755.5 torr – 17.54 torr

                                      = 737.96 torr

b. Calculate the moles of nitrogen present with 425ml of N2 are collected over water at 20°C and 755.5 torr.

Solution :- volume of N2 = 425 ml = 0.425 L

Temperature T= 20 C +273 = 293 K

Pressure of N2 = 737.96 torr * 1 atm / 760 torr =0.971 atm

Moles of N2 = ?

Using the ideal gas law formula we can calculate the moles of the N2

PV= nRT

Where P= pressure , V= volume , R= gas constant (0.08206 L atm per K mol) and

T= temperature

Lets put the values in the formula

0.971 atm * 0.425 L = n *0.08206 L atm per K mol* 293 K

(0.971 atm * 0.425 L) /(0.08206 L atm per K mol* 293 K) = n

0.0171636 mol N2 = n

Therefore moles of N2 = 0.0171636 mol

c. How many moles of HSO3NH2 would have to react to produce the number of moles of nitrogen calculated in problem b? How many grams of HSO3NH4?

Solution :- mole ratio of the N2 to HSO3NH4 is 1 :1 therefore the moles of the HSO3NH4 are same as moles of N2

Therefore moles of the HSO3NH4 = 0.0171636 mol

d. Assume that the HSO3NH2 sample weighed 3.125g. Calculate the percent purity of the sulfamic acid.

Solution :-

Lets convert moles of the HSO3NH4 to the mass

Mass = moles * molar mass

         = 0.0171636 mol * 99.1096 g per mol

        = 1.701 g HSO3NH4

Now lets calculate its percent purity

% purity = (actual mass of HSO3NH4 / mass of sample )*100 %

                = (1.701 g / 3.125 g)*100%

              = 54.43 %

Therefore the percent purity of the HSO3NH4 is 54.43 %

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