Question

Lab Name:

Analysis of an Unknown Mixture Using the Ideal Gas Law

Purporse:

Determine the percent composition by mass of an unknownmixture of sodium nitrite, NaNO2, and sodium chlordie, NaCl, aftercollecting a gas evolved by reaction of NaNO2.

Pre-Lab Notes:

In this experiment you will react a mixture of sodium nitrite,NaNO2, and sodium chloride, NaCl, of unknown composition with anexcess of sulfamic acid, HSO3NH2. The NaCl is unreactive underthese conditions whereas the NaNO2 reacts to form nitrogen gas byfollowing equation."

NO2(aq) + HSO3NH2(aq) --> HSO4(aq) + H2O(l) + N2(g).

You will collect the N2 gas at atmospheric pressure and roomtemperature and calculate from your data the moles of N2, moles ofNaNO2, mass of NaNo2, mass of NaCl and percent composition by massof NaNO2 and NaCl.

Lab Data:

Masses, Volumes, temperatures, and pressure

Mass of vial (grams) - 3.6450

Mass of vial and unknown sample (grams) - 3.7450

Initial volume of gas (mL) - 10.1

Final volume of gas (mL) - 27.6

Barometric pressure (mmHg) - 755.5

Room Temperature ( C ) - 23.5 C degress

Calculations required

Mass of unknown sample (grams) - 0.1

A) Volume of N2 gas collected ( mL) - 17.5mL

B) Vapor pressure of water at room temperature(mmHg) - 21.75mmHg

C) Pressure of dry N2 gas (mmHg) -

D) Moles of N2 collected (mol) -

E) Moles of NaNO2 reacted (mol) -

F) Mass of NaNO2 reacted (grams) -

G) Percent NaNO2 by mass in unknown sample(%) -

H) Average percent NaNO2 by mass in unknown sample(%) -

I) Mass of NaCl in unknown sample (grams) -

J) Percent of NaCl by mass in unknown sample (%) -

K) Average percent of NaCl by mass in unknown sample (%) -

please can you leave the solution and the calculation? i have no clue.. please help I would sincerely appreciate it!

Answer 1

A) Volume of N2 gas collected ( mL) - 17.5mL

B) Vapor pressure of water at room temperature(mmHg) - 21.75mmHg

C) Pressure of dry N2 gas (mmHg) –

=Total pressure - vapor pressure of water

= 755.5 mmHg- 21.75 mmHg

= 733.75 mmHg

= 0.965 atm

D) Moles of N2 collected (mol) –

PV= nRT

n = PV/RT

n= 0.965 atm*0.0175 L*0.08206 atm-L/mol –K*296.5 K

n= 6.94*10^-4 mol

E) Moles of NaNO2 reacted (mol) –

NO2(aq) + HSO3NH2(aq) --> HSO4(aq) + H2O(l) + N2(g).

From reaction we can see that 1 mol of N2 produce from 1 mol of NO2 thus the moles of NO2 =

6.94*10^-4 mol

1 mole of NO2 comes form 1 mole of NaNO2

F) Mass of NaNO2 reacted (grams) –

Number of moles * molar mass

=6.94*10^-4 mol*68.9953 g/mol

=0.048 g

G) Percent NaNO2 by mass in unknown sample(%) –

Mass of unknown sample (grams) - 0.1

Percent NaNO2 by mass in unknown sample(%)= mass of NaNO2 / total mass of sample*100

=0.048 g/0.1 g*100

=48%