What is the limiting reactant when 3 grams of propane is mixed with 10 grams of Oxygen?

C3H8 + O2 —> H2O + CO2

The answer to your question is Propane

Explanation:

Data

mass of Propane = 3 g

mass of Oxygen = 10 g

limiting reactant = ?

Balanced chemical reaction

               C₃H₈  +  O₂  ⇒   H₂O  +  CO₂

Process

1.- Calculate the molar mass of the reactants

C₃H₈ = (12 x 3) + (8 x 1) = 36 + 8 = 44 g

O₂ = 16 x 2 = 32 g

2.- Calculate the theoretical yield  C₃H₈ / O₂ = 44/32 = 1.375

3.- Calculate the experimental yield C₃H₈/O₂ = 3/10 = 0.3

4.- Conclusion

The limiting reactant is Propane because the experimental yield was lower than the theoretical yield.

explanation:

let the different atoms be x and y. each with four protons and same mass will be represented as follows.

mass number is number of protons + number of neutrons.

            and  

since, it is given that one has a negative charge while the other is neutral. therefore, there will be increase in number of protons and the atom will be represented as follows.

according to periodic table, mass number of beryllium is 9 and it has atomic number 4 so, the atoms will be and .  

the protons and neutrons will be located inside the nucleus whereas the electrons will be revolving around the nucleus.  

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