Question

Consider the balanced reaction below. How many moles of oxygen would have be consumed to produce 68.1 g water? C3H8+5O2--->3CO2+4H2O

Don't understand any of this, if you could please include a step by step... thank you so much!

Answer 1

The answer is....`0.945*mol` with respect to dioxygen gas. And note that I work from the given equation.....

Explanation:

Look at your stoichiometrically balanced equation: for every REACTANT ATOM there is necessarily a PRODUCT ATOM:

`C_3H_8(g) + 5O_2(g)rarr3CO_2(g) + 4H_2O(l) + Delta`

And since atoms have definite, and measurable masses, it necessarily follows that the mass of the reactants is PRECISELY equal to the mass of the products. And what does `Delta` represent?

This same principle of stoichiometry, of balance, is observed in banking, and finance, and even a simple purchase from a shop. The value of the purchase PLUS the change you receive from the merchant, is EQUAL to the value of the bill you gave the merchant. The debit on your bank account MUST equal the credit made to the shopkeeper's account.....

Now with respect to water we get a molar quantity of `(68.1*g)/(18.01*g*mol^-1)=3.78*mol` (from where did I get `18.01*g*mol^-1`?). Now given the equation (and we assume quantitative yield in that we have no way of knowing otherwise), there were PRECISELY `(3.78*mol)/4` with respect to pentane, i.e. a molar quantity of `0.945*mol`, and a mass of `68.2*g`; and again I work from the equation. And we thus require five equiv of dioxygen gas..... a molar quantity `4.73*mol` and a mass of `151.2*g`.

Are you happy with this? It is an important principle to understand.