C6H14+9.5O2=6CO2 +7H20
Number of moles of C6H14=15.6/86=0.1814 moles
so moles of CO2 = 6(0.1814)=1.088
As the c6h14 has 1 is to 6 ratio with co2
so
0.1814=mass/44
mass of co2 produced = 47.9 g
<span>The rate of effusion of a gas is inversely proportional to the square root of the molecular weight of the species. Now there will be differences among isotopomers but neglecting these and taking the avg mol wt of N2 = 28 and Xe = 132;
Rate(N2)/Rate(Xe) = sqrt (132/28) = 2.17</span>
The volume of the gas at STP = 35.01 L
<h3>Further explanation</h3>
Conditions at T 0 ° C and P 1 atm are stated by STP (Standard Temperature and Pressure).
In general, the gas equation can be written

where
P = pressure, atm
V = volume, liter
n = number of moles
R = gas constant = 0.08206 L.atm / mol K
T = temperature, Kelvin
V=17.4 L
T = 23 + 273 = 296 K
P = 2.18 atm

The volume of the gas occupy at STP :

<span>134 ml
First, let's determine how many moles of oxygen we have.
Atomic weight oxygen = 15.999
Molar mass O2 = 2*15.999 = 31.998 g/mol
We have 3 drops at 0.050 ml each for a total volume of 3*0.050ml = 0.150 ml
Since the density is 1.149 g/mol, we have 1.149 g/ml * 0.150 ml = 0.17235 g of O2
Divide the number of grams by the molar mass to get the number of moles
0.17235 g / 31.998 g/mol = 0.005386274 mol
Now we can use the ideal gas law. The equation
PV = nRT
where
P = pressure (1.0 atm)
V = volume
n = number of moles (0.005386274 mol)
R = ideal gas constant (0.082057338 L*atm/(K*mol) )
T = Absolute temperature ( 30 + 273.15 = 303.15 K)
Now take the formula and solve for V, then substitute the known values and solve.
PV = nRT
V = nRT/P
V = 0.005386274 mol * 0.082057338 L*atm/(K*mol) * 303.15 K / 1.0 atm
V = 0.000441983 L*atm/(K*) * 303.15 K / 1.0 atm
V = 0.133987239 L*atm / 1.0 atm
V = 0.133987239 L
So the volume (rounded to 3 significant figures) will be 134 ml.</span>