Hi. I just took a quiz and I totally skipped this Question because I do not understand at all. Can someone Explain??

Question

anonymous · Answer

If enough oxygen is present for 132 of propane (C3H8) to burn completely in the given  reaction, what mass of water is produced?

anonymous · Answer

..............................Is there no one there...
If not I need to go really soon....

photon336 · Answer

$$C_{3}H_{8} + 5O_{2} -->3CO_{2} + 4H_{2}O$$

photon336 · Answer

you need the molar mass of propane this is done in excess oxygen, that's what complete combustion is. 

we know that our limiting reagent will be the propane. so we find out how many moles are in propane.

$$132g(C_{3}H_{8})*(\frac{ mol C_{3}H_{8} }{ 44.1g }) = \frac{ 132 mol }{ 44.1} = 3 mol C_{3}H_{8}$$

then we multiply by the molar ratio to find the moles of H2O notice something? moles of C3H8 cancel out and we're left with moles of H2O

$$3 mol, C_{3}H_{8}*(\frac{ 4H_{2}O }{ C_{3}H_{8} }) = 12 mol H_{2}O$$

now we need the mass of H2O we do this by doing the following below:

$$\frac{ 18.0 g}{ mol }*12mol = 216grams, H_{2}O$$