1) A gas mixture is made by combining 6.9 g each of Ar, Ne, and an unknown diato
ID: 996346 • Letter: 1
Question
1) A gas mixture is made by combining 6.9 g each of Ar, Ne, and an unknown diatomic gas. At STP, the mixture occupies a volume of 88.20 L.
- What is the molar mass of the unknown gas?
_______ g/mol
- Identify the unknown gas
( F2, Cl2, O2, H2, N2)
2) Container A holds 777 mL of ideal gas at 2.70 atm. Container B holds 129 mL of ideak gas at 4.90 atm. If the gases are allowed to mix together, what is the resulting pressure?
- ______ atm
3) If He(g) has an average kinetic energy of 8890 J/mol under certain conditions, what is the root mean square speed of F2(g) molecules under the same conditions?
- _________ m/s
4) Chlorine can be prepared in the laboratory by the reaction of manganese dioxide with hydrochloric acid, HCl (aq), as described by the chemical equation
MnO2(s)+ 4HCl(aq) MnCl2(aq)+2H2O(l)+ Cl2(g)
-How much MnO2(s) should be added to excess HCl(aq) to obtain 155 mL of Cl2(g) at 25°C and 755 Torr?
_______g MnO2
Explanation / Answer
1.
Molar mass : Ar= 40 , Ne=20
Moles = mass/Molecular weight
Moles : Ar= 6.9/40=0.1725, Ne= 6.9/20 =0.3450
Let the other gas moles be =x
Total moles of mixture = 0.1725+0.3450+x=0.5175+x
22.4 L of gas correspond to 1 mole of gas
88.2 L of gas correspond to 88.2/22.4=3.9375 moles
But 0.5175+x= 3.9375
Therefore x= 3.9375-0.5175=3.42 moles
Moles =mass/Molecular weight
3.42= 6.9/Molecular weight
Molecular weight= 6.9/3.42=2
The gas is H2
2.
Moles of gas in container A =PV/RT= 2.7*0.777/RT= 2.1/RT
Moles of gas in container B= 0.129*4.9/RT= 0.6321/RT
Total moles = (2.1+0.6321)/RT=2.7321/RT
Volume after mixing = 777+127ml = 904ml=0.904L
From PV= nRT, the resulting pressure
P = nRT/V= 2.7321/0.904 =3.02 atm
3. Molecular weight of Fe= 38 g/mole
kinetic energy= (1/2)mV2)= 0.5*38/1000*V2= 8890
V= 684 m/s
4.
Moles of chlorine = PV/RT= (755/760)*0.155/(0.0821*(25+273.15)=0.0063 moles
From the reaction 0.0063 mole of chlorine is produced from 0.0063 mole of MnO2
Moles of MnO2=0.0063
Molecular weight of MnO2=87, mass of MnO2 required= 87*0.0063=0.5481 gms
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