A sample of methane gas (CH4 (g)) having a volume of 2.8 L at 25oC and 1.65 atm was mixed with a sample of oxygen gas having a volume of 35 L at 31oC and 1.25 atm. The mixture was then ignited to form carbon dioxide and water. Calculate the volume of CO2 formed at a pressure of 2.5 atm and a temperature of 125oC.

2.47 L

Explanation:

The first thing we do is to write the balanced equation, which is.

CH4(g) + 2O2(g) --> CO2(g) + 2H2O(g)

Now, using ideal gas equation, we see that

PV = nRT

moles of methane

where P is the pressure, 1.65 atm

V is the volume, 2.8 L

R is the ideal gas constant, 0.0821

T is the temperature in Kelvin, 25 + 273 = 298 K

n = PV/RT

n of methane = 1.65 * 2.8 / 0.0821 * 298

n of methane = 4.62 / 24.47

n of methane = 0.189 mole

moles of O2:

PV = nRT

where P is the pressure 1.25 atm

V is the volume, 35 L

R is the ideal gas constant, 0.0821 L

T is the temperature, 31 + 273 = 304 K

n of O2 = 1.25 * 35 / 0.0821 * 304

n of O2 = 43.75 / 24.96

n of O2 = 1.75 mol

from the equation, the molar ratio between O2 and CH4 is 2:1, so therefore

no of moles of O2 = 2 * .0189 moles of methane = 0.378 mole

when 1 mole of CH4 --> 1 mole of CO2

1 mol of CO2 --> 22.4 L CO2

volume of CO2 at STP = [(0.189 moles of CH4 /1) * (1 mol of CO2)] / [(1 mole CH4) * (22.4L CO2)/(1 mol of CO2)] = 4.23 L

Next, we use the formula

P1V1 /T1 = P2V2/T2

where

P1 = 1 atm

V1 = 4.23 L

T1 = 273

P2 = 2.5 atm

T2 = 125 + 273 = 398 K, making V2 subject of formula, we have

V2 = P1V1T2 / T1P2

V2 = 1 atm * 4.23 L * 398 / 273 * 2.5 atm

V2 = 1683.54 / 682.5

V2 = 2.47 L

"you start with 15.67 g mgco3-xh2o, when heated and all the water is removed, you are left with 7.58 g mgco3 only.

therefore, you had 15.67 - 7.58 = 8.09g h2o  

moles of each :

7.58g mgco3 = ? moles = 7.58g / 83.4g/mole = 0.09moles  

8 09g h2o = ? moles = 8.09g / 18g/mole = 0.44moles  

molar ratio between the two by dividing both numbers by the smallest number :

0.09 / 0.09 = 1  

0.44 / 0.09 = 4.8  

mgco3-5h2o"