1. The following equation represents the partial combustion of methane, CH 4 . 2

Question

1.

The following equation represents the partial combustion of methane, CH4.

2CH4(g) + 3O2(g)    2CO(g) + 4H2O(g)

At constant temperature and pressure, what is the maximum volume of carbon monoxide that can be obtained from 2.12 × 102 L of methane and 1.06 × 102 L of oxygen?

A.4.24 × 102 L

B.7.42 × 102 L

C.2.12 × 102 L

D.3.18 × 102 L

E.7.07 × 101 L

2.

The density of ethane, C2H6 (30.1 g/mol), at 22°C and 1.01 atm pressure is:

A.1.26 g/L

B.0.797 g/L

C.14.8 g/L

D.1.34 g/L

E.0.121 g/L

3.

When the valve between the 2.00-L bulb, in which the gas pressure is 1.00 atm, and the 3.00-L bulb, in which the gas pressure is 3.00 atm, is opened, what will be the final pressure in the two bulbs? Assume the temperature remains constant.

A.1.67 atm

B.2.20 atm

C.1.80 atm

D.4.00 atm

E.2.33 atm

4.

What is the net ionic equation for the reaction of NH3 with HNO3?

A.H–(aq) + H+(aq) H2(g)

B.NH3(aq) + H+(aq) NH4+(aq)

C.H+(aq) + OH–(aq) H2O(l)

D.NH3(aq) + HNO3(aq) NH4NO3(aq)

E.NH3(aq) + NO3–(aq) NH2–(aq) + HNO3(aq)

Explanation / Answer

1 ) answer : E ) .7.07 × 101 L

solution : 2CH4(g) + 3O2(g)    2CO(g) + 4H2O(g)

at STP condition any 1 mol gas can occupy 22.4 L

2 mole CH4 volume = 22.4 x 2 = 44.8 L

3 mol O2 volume = 22.4 x 3 = 67.2 L

2 mole CO volume = 22.4 x 2 = 44.8 L

2CH4(g) +    3O2(g) -----------------> 2CO(g) + 4H2O(g)

44.8 L              67.2 L                            44.8L

2.12 x 10^2L     1.06 x 10^2 L                 ?

here limiting reagent is O2

so

67.2 L O2 ------------------> 44.8 L CO

1.06 x 10^2 L --------------> ?

CO volume = 44.8 x 1.06 x 10^2 / 67.2

                     = 70.7 L

                     = 7.07 x 101 L

2) answer : A) .1.26 g/L

solution :

pressure (P) = 1.01 atm

temperature (T) = 273 + 22= 295K

molar mass of C2H6 (M) = 30.1 g/ mol

R = universal gas constant = 0.0821 L-atm / K mol

ideal gas equation

P V = n RT

PV = (w/M)RT

P = (w/V)RT/ M

P = dRT / M

d = P M / RT

    = 1.01 x 30.1 / 0.0821 x 295

     = 1.26 g / L

3) answer : A ) 1.67 atm

solution :

P1 = 1 , V1 = 2

P2 = ? , V2 = 3

P1 V1 = P2 V2

P2 = P1 V1 / V2 = 1 x 2/3 = 0.67

total pressure in two bulbs = P1 + P2 = 1 + 0.67 = 1.67 atm

4) answer : B) .NH3(aq) + H+(aq) NH4+(aq)

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