2 NaN,(s) 2 Na (s) 3 Na (g) How many grams of sodium azide are needed to provide

Question

2 NaN,(s) 2 Na (s) 3 Na (g) How many grams of sodium azide are needed to provide sufficient nitrogen gas to fill a 30.0 x30.0 x 25.0 cm bag to a pressure of 1.20 atm at 1.0 x10%C? Number g sodium azide 4 Koz(s) 2 cozle) 2 K co,(s) 3 oz(g) How many grams of KO2 are needed to produce 1125.0 L of O2 at 20.0°C and 1.00 atm? Number A sample of a gas weighs 2.518 g and has a volume of 1100.0 mL at 985.0 torr and 27.0°C. What is its molar mass? Number g mol A 4.00 L sample of gas is cooled from 71 CC to a temperature at which its volume is 2.60 L. What is this new temperature? Assume no change in pressure of the gas. Number 109.2

Explanation / Answer

Solution:

Given: Pressure nitrogen gas = 1.20 atm , T= 1.0 101 deg C = 10 deg C

Volume of bag = 30.0 cm * 30.0 cm * 25.0 cm =22500 cm3 = 22500 mL    ( 1mL = 1 cm3)

Conversion of temperature into K

Temperature in K = 10 deg C + 273.15 K = 283.15 K

Volume in L = 22500 mL * 1 L / 1000 mL = 22.500 L

By using given date and ideal gas law we find moles of N2

Ideal gas law :

PV = nRT

R = gas constant = 0.08206 L atm / K mol

n = PV/ RT = 1.20 atm * 22.500 L / (0.08206 L atm / K mol * 283.15 K )

= 1.16

Moles of N2 = 1.16 mol

Mole ratio of N2 : NaN3 is 3 : 2

We use this mole ratio to get moles of NaN3 from moles of N2 and by using molar mass we get mass of sodium azide

Moles of NaN3 = 1.16 mol N2 * 2 mol NaN3/ 3 mol N2

= 0.775 mol NaN3

Mass of NaN3 in gram = mol NaN3 * molar mass

= 0.775 mol NaN3 * 65.011 g/mol

= 50.36 g

Mass of sodium azide needed is 50.36 g

Q. 2.

Solution:

Given volume of O2 = 1125.0 L , T = 20.0 deg C , Pressure = 1.00 atm

Conversion of temperature to K

Temp in K = 20.0 deg C +273.15 =293.15 K

We calculate moles of O2

n = PV/ RT = 1.00 atm * 1125.0 L / ( 0.08206 L atm / K mol * 293.15 K )

= 46.77 mol

We use mole ratio to calculate moles of KO2

Mole ratio of O2 : KO2 is 3:1

Mol KO2 = mol O2 * 1 mol KO2 / 3 mol O2

= 46.77 mol O2 * 1mol KO2 / 3 mol O2

= 15.59 mol KO2

Mass of KO2 in g = moles of KO2 * molar mass

= 15.59 mol KO2 * 71.096 g /mol

=1108.29 g KO2

The mass of KO2 needed is 1108.29 g

Q. 3

Solution :

Given mass of gas = 2.518 g

Volume = 1100.0 mL

Pressure = 985.0 torr

T = 27.0 deg C = 27.0 deg C+ 273.15 K = 300.15 K

Lets convert pressure into atm

1 atm = 760 torr

So atm in 985.0 torr = 985.0 torr * 1atm / 760 torr = 1.30 atm

We use Ideal gas law and some rearrangement in that .

pV= nRT

we know n = m/M , here m is mass in gram and M is molar mass

we plug value in n in above equation

pV = m/ M (RT )

Lets rearrange the equation so that we get M to the left side

M = m RT / PV

Let plug given value in above equation

M = [2.518 g *( 0.08206 L atm/ K mol) * 300.15 K] / ( 1.30 atm * 1100.00 L )

= 6.44 g/mol

Q. 4

Given : Initial Volume of gas = 4.00 L ,Initial T= 71 deg C

Final volume = 2.60 L , T2= Unknown

We use P1V1 T2 = P2V2 T1

Here P is constant so

V1T2 =V2T1

Lets convert T in K

T1 in K = 71 deg C +273.15 = 344.15 K

T2 = V2 * T1 / V1

= 2.60 L * 344.15 K / 4.00 L

= 223.70 K

T in deg C = 223.70 -273.15 = -49.45 deg C

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