How many molecules of co2,h2o, c3h8, and o2 will be present if the reaction goes to completion?

Let initially there are 10 molecules of O2 and 3 molecules of C3H8 present

The reaction will be

C3H8(g)  + 5O2(g)  > 3CO2(g)   + 4H2O

so here oxygen molecules are limiting as for 3 molecules of C3H8 we need 15 molecules of O2

now the given 10 molecules of O2 will react with only 2 molecules of C3H8 and they will form six molecules of CO2 and 8 molecules of H2O

Hence answer is

molecules of CO2 formed = 6

Molecules of H2O formed = 8

molecules of C3H8 left = 1

molecules of O2 left = 0

Molecules of CO₂, H₂O, C3H8, and O₂ will be present if the reaction goes to completion are  

0 molecules C3H8

0 O₂ molecules

3 and CO₂ molecules

4 H₂O molecules

The reaction equation is the chemical formula of reagents and product substances

The reaction coefficient is a number in the chemical formula of a substance involved in the reaction equation. The reaction coefficient is useful for equalizing reagents and products.

Steps in equalizing the reaction equation:

•1. gives a coefficient on substances involved in the equation of reaction such as a, b, or c, etc.

•2. make an equation based on the similarity of the number of atoms where the number of atoms = coefficient × index between reactant and product

•3. Select the coefficient of the substance with the most complex chemical formula equal to 1

The reaction of the combustion of hydrocarbons with oxygen produces CO₂ and H₂O (water vapor). can use steps:

Balancing C atoms, H and last atoms O atoms

Equivalent reaction equation, if the number of atomic substances of the reactor is the same as the product substance

The reaction coefficient also shows the ratio of moles, volumes or molecules of substances involved in the reaction

Complete combustion of Hydrocarbons with Oxygen will be obtained by CO₂ and H₂O compounds.

If O₂ is insufficient there will be incomplete combustion resulting in CO and H₂O

In perfect combustion of propane(C3H8) and oxygen (O₂), propane and oxygen will be converted entirely into CO₂ and H₂O at the end of the reaction

Propane and oxygen reactions:

C₃H₈ + O₂ ---> CO₂ + H₂O

We equalize first by giving coefficients a, b and c

C₃H₈ + aO₂ ---> bCO₂ + cH₂O

C = left 3, right b -> b = 3

H = left 8, right 2c ---> c = 4

O = left 2a , right 2b + c

2a =  2b+c

2a = 2 . 3 + 4

2a = 10

a = 5

So the equivalent reaction equation =

C₃H₈ + 5O₂ ---> 3CO₂ + 4H₂O

The comparison of the coefficients shows that: 1 C3H8 molecule reacts with 5 molecule O₂ produces 3 molecules of CO₂ and 4 molecules of H₂O

So the complete combustion of ethanol at the end of the reaction composition:

0 molecules C3H8

0 O₂ molecules

3 CO₂ molecules

4 H₂O molecules

the combustion of ethanol

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synthesize 18.0 mol of no

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mass of copper required

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Keywords : combustion, propane, molecules

the ideal gas law equation is as follows

pv = nrt

where p - pressure

v - volume

n - number of moles

r - universal gas constant

t - temperature

so if the pressure, volume and temperature are already known

we are left with n and r

since r is the universal gas constant that has a known fixed value then r too is known

so we are left with 'n'

once we know temperature volume and pressure

we can find the number of moles of gas present using the ideal gas law equation

this is not my own answer, this is an answer for the same question by the user blahblahmali. i do not claim ownership of this account.

i think that the answer is c