What mass of o_2 is required to produce 5.00 l no_2 at 305 k and 752 mm hg?(r =

Question

What mass of o_2 is required to produce 5.00 l no_2 at 305 k and 752 mm hg?(r = 0.08206 l atm/mol K)2 no(g) + O_2(g) rightarrow 2 NO_2(g) 0.198 g 1.58 g 3.16 g 5.00 g 6.33 g Equal volumes of propane, C_3H_8. and carbon monoxide. CO. at the same temperature and have the same chemical properties. number of atoms. average molecular speed. density. number of molecules. The pressure of O_2 in a 15.0 flask is 322 mm Hg at 44 degreeC What mass of O_2 is in the flask?(R = 0.0K06 L atm/mol K) 0.244 g 5.80 g 7.82 g 15.2 g c. 56.3 g

Explanation / Answer

41)Using ideal gas equation, moles of NO2 required can be calculated.

PV=nRT

n=PV/RT

P=752/760 =0.99atm

1 atm=760 mm hg

V=5.00L

R=0.0821L atm/K.mol

n=0.99 atm * 5.00L/(0.0821L atm/K.mol)*305K=0.198 moles

From the balanced equation you can see the coefficients and figure out ,

1 mol O2 is required to produce 2 mol NO2

So moles of O2 reqd to produce 0.198 moles NO2=1/2*0.198=0.099=0.1 moles

1 mole O2=32 g

0.1 mol O2=0.1*32=3.2 g(answer)

C) is correct

42)using Boyle’s law ,

P1*V1=P2*V2

P1=1.6 atm

V1=45L

P2=0.7 atm

V2=?

V2=P1*V1/P2=1.6atm*45L/0.7 atm=103 L e) correct

43)e )No of molecules.

Avogadro’s law states that equal volume of all gases at equal P,T has same no of molecules equal to 6.022*10^23(Avogadro’s number)

44) Using ideal gas equation, moles of O2 required can be calculated.

PV=nRT

n=PV/RT

P=322/760=0.42 atm

T=44+273=317 K

V=15.0L

R= R=0.0821L atm/K.mol

n=0.42 atm * 15.0L/( 0.0821L atm/K.mol)* 317K=0.242 moles

1 mol O2=32 g

0.0242 moles O2=0.0242 moles*32g/mol=7.8 g(answer)

c) is correct

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