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

1 Answer
May 24, 2017

Answer:

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?