Question

How many moles of water can be produced from the reaction of 28 g of `C_3H_8`?

Answer 1

I presume you speak of a combustion reaction? All hydrocarbons combust completely to give carbon dioxide and water.........Here we gets approx. `2.5*mol`.

Explanation:

First we need a stoichiometric equation, and this is something at which you need to be very adept. The usual rigmarole is: balance the `"carbons (as carbon dioxide)"`; then balance the `"hydrogens (as water)"`; then balance the `"oxygens"`.

`C_3H_8(g) +O_2(g)rarr CO_2(g) + H_2O(l)`

`"Balance the carbons.........."`

`C_3H_8(g) +O_2(g)rarr 3CO_2(g) + H_2O(l)`

`"Balance the hydrogens.........."`

`C_3H_8(g) +O_2(g)rarr 3CO_2(g) + 4H_2O(l)`

`"Balance the oxygens.........."`

`C_3H_8(g) +5O_2(g)rarr 3CO_2(g) + 4H_2O(l)`

With an even-numbered alkane, we would need a half-integral coefficient; try it out for butane, `C_4H_10`.

Back to your question..........we has............

`C_3H_8(g) +5O_2(g)rarr 3CO_2(g) + 4H_2O(l)`

We know that `1*mol` propane gives `3 *mol` carbon dioxide, and `4*mol` of water......In this reaction, we combusted a molar quantity of...................

`"Moles of propane"=(28.0*g)/(44.1*g*mol^-1)=0.635*mol`.

And given the stoichiometry of the reaction, the water that arises from the combustion is `0.635*molxx4=??`

What mass of water does this molar quantity represent?