What volume of O2 is required to completely burn 15cm3 of pentane? Equation - C5H12 + O2 - CO2 + H2O.​

The volume of required would be , assuming that both the pentane and the in this question are ideal gases and are under the same temperature and pressure.

Explanation:

Balance the equation for the reaction:

.

Start by setting the coefficient of the molecule with the largest number of atoms to .

In the combustion of alkanes (including pentane,) consider setting the coefficient of the alkane to .

.

Number of carbon atoms among the reactants: .

Number of hydrogen atoms among the products: .

By the conservation of atoms, there would need to be the same number of carbon and hydrogen atoms (along with the oxygen atoms) among the products.

Hence, the coefficient of would be while the coefficient of would be .

.

There would be oxygen atoms among the products. Also by the conservation of atoms, there would be the same number of oxygen atoms among the reactants.

Hence, the coefficient of would be .

.

The ratio between the coefficient of and in the balanced equation is:

.

In other words, it would take eight molecules to react with one molecule.

Assume that both and are ideal gases. Under the same temperature and pressure, the volume of the two gases would be proportional to the number of molecules in each gas:

.

In other words, it would take of to react with of under these assumptions. It would then take of to react with of .

a: 45.9 meters

b: 4.29 seconds

step-by-step explanation:

a: h(t)=1/2(-9.8)(3)^2+0(3)+90

h(t)=-4.9(3)^2+90

h(t)=-44.1+90

h(t)=45.9 meters

b: 0=1/2(-9.8)(t^2)+0t+90

0=-4.9t^2+90

-90=-4.9t^2

18.37=t^2

t=4.29 seconds

in the second question, the numbers aren't exact, but they're a close approximation.